Hyunwoo Kwon (Will)

Sarah Helen Whitman, who saved Edgar Allen Poe

2026-03-08T10:00:40+00:00

An original version is Korean and is given in the link. The following translation is based on Claude with some minor modification.

Today, March 8th, is International Women’s Day. It originated in demonstration of the tens of thousands of textile workers who took to the streets of New York in 1908, though why March 8th specifically became the designated date remains unclear. It’s a history less than 120 years old, yet even the origins of how this day was established aren’t widely known.

This is a story from my time dating someone in Providence, someone I’m no longer in contact with. She loved horror movies, and she recommended Netflix’s “The Fall of the House of Usher” to me. I later discovered the story was written by Edgar Allan Poe, and it wasn’t until a year had passed that I learned how deeply connected Poe was to Providence.

After moving into my current apartment, I found that the library near my house was one where Edgar Allan Poe had once lectured, and it was the first time that I heard the name Sarah Helen Whitman as Poe’s lover. In her time, Whitman was a well-regarded literary critic, while Poe was a writer only beginning to gain recognition across America through his poem “The Raven”, just a year before his death. The two became engaged, but various circumstances led to the engagement being broken off, and Edgar died in the street a year later.

Poe did not achieve popular success until he was 36. American literature at the time was dominated by transcendentalism — figures like Ralph Waldo Emerson — and by Puritan-centered writing. His novels written around his thirties, “Ligeia” and “The Fall of the House of Usher”, went largely unnoticed in their day, and he could not sustain himself financially through fiction alone. But once “The Raven” brought him success, he began receiving invitations to speak at literary salons.

Frances Osgood, who had frequent exchanges with Poe, invited him to Providence. She intended to introduce him to Whitman, already a celebrated writer and literary critic in New England, but as the story goes, Edgar caught sight of Whitman standing in a rose garden from a distance and could not bring himself to speak to her. Yet Whitman had been reading his work four or five years before he became famous, and as a tribute to him, she wrote a Valentine’s Day poem praising his artistic sensibility.

Whitman was born and raised in Providence, and was the prominent literary figures of literature circles in New England in her era. She was close with Ralph Waldo Emerson and described herself as his disciple, so great was his influence on her. She corresponded with Margaret Fuller, who wrote what is considered one of the first works of feminist literature, and served as vice president of the Rhode Island Woman Suffrage Association.

She contributed more than 80 essays to newspapers. In 1849, Rufus Griswold, the most influential literary editor of the day, included Whitman as a representative American female poet in his anthology, even singling her out in his preface. Horace Greeley, founder of the New York Tribune, also sought her contributions, leaving no doubt that she was a prominent literary figure of her time.

When Edgar Allan Poe died in 1849, Griswold published an obituary in the newspaper under the pseudonym “Ludwig,” portraying Poe as a drunkard, a fraud, and a man of ruined character. Griswold had been designated as the editor of Poe’s posthumous works, ironically, by Poe himself, and the preface to Poe’s collected works contained a deeply distorted biography. From that point on, Whitman wrote a book in defense of Poe’s work, and after Griswold’s own death, she produced a masterful piece logically refuting all charges against Poe’s literary merit. Yet 160 years later, Whitman is remembered only as Edgar Poe’s former fiancée. What is spoken of are the letters they exchanged, their fateful meeting, and the turbulent dissolution of their engagement, while the 17 years she spent working to restore Poe’s reputation within American literary circles go largely unnoticed.

Edgar Poe and His Critics, written in 1860, reads even today as a work of remarkable logical precision, one that hardly feels 160 years old. It directly refutes Griswold’s 1849 exposé as a text filled with distorted facts and malicious assumptions. To accomplish this, Whitman gathered information from contemporaries who had known Poe, added her own critical insights, and argued methodically for why Poe’s work deserved serious attention.

Critics of the time praised his facility with language but refused to recognize the artistic merit of his work, unsurprising given that the prevailing moralism of Puritan culture made it nearly impossible for gothic horror fiction to find its light. In this essay, Whitman articulates the value of such literature as “a product of deep inner experience and the projection of intense imagination,” writing with measured scholarly calm rooted in her personal memories. She squarely refutes the charge that Poe lacked moral sensibility by drawing on his past work as evidence, and writes with quiet empathy about the drinking that had led to their broken engagement, approaching it from his perspective.

In writing this, Whitman referred to herself as “one of the friends who remembers him,” crafting a precise and deliberate piece so that future generations might judge Poe’s genius fairly. The essay is also regarded as the first work to logically separate an author’s private life from the artistic integrity of their work, a distinction made before Oscar Wilde, and a full 80 years before Roland Barthes argued that a work’s meaning is completed in the moment a reader encounters it, regardless of the author’s intent. It is a distinguished piece of writing that steps beyond the “biographical criticism” that dominated the era and argues, with clear logic, why Poe should not be undervalued.

In 1873, British John Ingram, outraged by Griswold’s false record, began a project to uncover the truth. As Ingram worked on a biography of Poe, he corresponded with Whitman. She provided him with Poe’s original letters and rare materials, and over five years they exchanged correspondence in which she explained and critiqued Poe’s literary theories and the circumstances of his life. Later, when Ingram realized that Whitman also corresponded with another biographer, he published another private letters as a news item, portraying Whitman in inappropriate way. She is said to have responded with a graceful critique of Ingram in the Providence Journal. The 75-year-old woman died of heart disease, complications, and ether addiction one month after writing that piece. Ingram’s biography would go on to establish Poe’s reputation in Europe as “America’s cursed genius.” Yet Whitman, whose contributions were among the greatest to that biography, is mentioned only as “Poe’s former fiancée.” The University of Virginia later commented that the collected letters contain “sharp and insightful discussions of Poe that biographers have not sufficiently utilized.” Mail correspondence can be seen in this link.

The moment when Poe could not bring himself to speak to Whitman tends to be flattened into a simple narrative, which he didn’t approach her because she was beautiful. It is true that Poe later wrote her love letters, but at that time he was a writer only belatedly beginning to receive attention, while Whitman was already a central figure in New England’s literary social world. Yet Ingram referred to the remarkable Edgar Poe and His Critics, the very essay that made his biography of Poe possible, merely as “a charming little book.”

I believe this story deserves to be more widely known. To read Whitman’s 17 years of advocacy after Poe’s death as nothing more than an affectionate tribute to a last love is, I think, to diminish the full arc of her life. Seventeen years of struggle cannot be reduced to love alone. It seems to me that she used every resource available to her, as an intellectual, to defend what she believed was genuine literary value. Without Whitman’s efforts, that singular gothic fiction might well have been forgotten, which might lost to the moral solemnity of antebellum America, or swept away in the aftermath of the Civil War. I find myself wondering how this story could be shaped into something for wider audiences, and who would love it — and then I remember: today is International Women’s Day.

Some announcement on AI: saying from Feynman, Laozi, and Larson

2026-01-09T10:00:40+00:00

A Korean translation is given in the link. English is the original version. This note is a draft for announcement in the following semester class.

Due to recent advances in generative AI, it is highly tempting to use it to solve a homework problem. Of course, I understand that you might feel it is easy to achieve a better score with AI, requiring less effort, but you also feel you are letting your brain outsource, becoming less competent than before. I think this is one of the environmental challenges we face because TikTok or YouTube Shorts weaken your concentration over time, and the weakened middle education due to COVID-19. I am also highly influenced by YouTube Shorts and Instagram Reels these days, and it is not that easy to deviate from them! I also understand that it is hard to concentrate on something more than before because there are so many tempting things that you could easily deviate from the path that you had to walk.

However, as a member of the teaching staff, it is my responsibility to help you get out of these trends and become who you want to be. As I talk with other students, this is not my own thinking, but you keep thinking about those concerns. I also feel that way after I use generative AI these days. I also actively use those when I want to sharpen my logic or seek some initial background, but I always double-check by reading some concrete references.

After ChatGPT launched, many students turned to generative AI to help them solve homework. It is totally okay to use generative AI to help your learning process, but it is not desirable to rely on those tools by letting your brain do the work for those computers. Your brain deserves to work. You should be aware of yourself, what you know, and what you don’t know. The point here is that active learning is more important than before.

I want to bring a famous quote from Richard Feynman: “The first principle is that you must not fool yourself, and you are the easiest person to fool. It is highly possible that you could say something that you don’t know. I found those in the advent of AI and in YouTube’s algorithm selection, which can lead to incorrect information or knowledge that is not easy to understand on that learning curve, and will give a misconception. Hence, it is more important than ever that you distinguish whether I really understand the key content or not. This lesson can also be found in an old saying from the Chinese philosopher Laozi: “To know that you do not know is the highest wisdom. To not know that you do not know is a disease. Only when one recognizes the disease as a disease can one be free from it”. In this age of AI, ‘pretending to know’ has become an easy trap to fall into, but it is a trap that prevents true growth.

In the last semester, I found that some students submitted homework they did not understand; for instance, imagine that elementary school students calculated the area of a triangle using calculus. This kind of logical fallacy makes it clear that the brain was not engaged. To protect the integrity of our learning community, I had to report those cases as academic misconduct. Please understand that our evaluations now depend heavily on in-person assessment; if you don’t understand the topic in your heart, you cannot pass or earn a satisfactory score if you rely solely on generative AI.

I know many of you have excelled in everything you have done so far. That’s why you’re here. Learning process is painful, and I am here to help you get through those difficulties. Failure will guide you to get a correct intuition to solve a problem, and discussion will lead to another big intellectual leap. I once spent three years struggling with a problem. I spent months stuck, feeling small as I watched others prove brilliant results. When I told my mentor, “It’s not working,” he gave me the most important lesson of my life. He said, “That’s good news. If it were easy, it wouldn’t be worth your time”. Failure wasn’t a waste of time. It was the process of sharpening my intuition. I still remember the moment of breakthrough. I was on a crowded metro when suddenly an abstract idea became a vivid image in my head, so I pulled out my notebook and began scribbling against the train door. That sudden flash of beauty is a human privilege that no AI can ever experience or give to you.

In the end, I want to ask you: what makes you distinct from generative AI? Many brilliant stories emerged from a unique idea rooted in life experiences. Frankenstein by Mary Shelley, The Fall of the House of Usher by Edgar Allan Poe, and Hamlet by Shakespeare (although the motivation is debated, the novel or movie Hamnet (2025) could offer an interesting theory) can be born from unique life experiences. Using generative AI alone will yield an average idea that is indistinguishable from any other. Mathematics is an art that captures phenomena in a rigorous, concise, and logical language. Great mathematics comes from a desire to solve or answer mysterious questions.

Throughout this course, I hope that you will learn something from mathematics that will help you answer those questions on your own. Jonathan Larson, who wrote the famous Broadway musical “Rent,” said, “The opposite of war is not peace; it’s creation.” Of course, his life was to struggle against his diseases as well as financial stuff. However, we are in a war for our attention, fighting against algorithms and the temptation of shortcuts. Being quiet or passive brings an illusion of peace, and actually, peace does not solely exist. The creation is achieved by actively engaging your brain to build something that belongs only to you. I hope that this mathematics course will give you a good experience after you leave Brown.

Art of Academic Writing: my perspective

2025-12-06T10:00:40+00:00

I wrote it in Korean first and I rewrote it in English.

I cannot remember where I saw the link (probably Facebook or LinkedIn). Someone made a toolkit for writing a paper that checks references and writing at the same time. While I agree that AI could greatly shape one’s point of view, I am highly skeptical of relying on generative AI to write a paper.

Writing a mathematics paper is a writing between poetry and a novel. It needs to be concise, like poetry, and explained well so the reader can understand the problem’s key difficulty. Although mathematics research paper is nonfiction, it needs a sort of elegance in elaborating the logic and it should contain some logical thinking so that people can be touched.

Many of the great papers that I read were in that category. For instance, the famous Landau damping paper by Mouhot and Villani clearly explains why this problem is important and elaborates on its key difficulties. Also, they explain the whole strategy for managing all difficulties. Another reason that I like this paper is that it appreciates people who discuss the problem with them. Creating a new language to explain the mysterious phenomenon can be achieved through fruitful discussions with people, not from a single brilliant mind. Moreover, this paper also explains potential different directions so that future researchers can get some benefit.

I also like the fundamental paper due to J. Leray written in 1934, Sur le mouvement d’un liquide visqueux emplissant l’espace. It explains many efforts by engineers, physicists, and mathematicians to understand the motion of viscous fluids. After Navier introduced the model, J. Leray sought to justify it by introducing a new framework. This is now called the theory of weak solutions. In this paper, he proved the global existence of turbulent solutions (now it is called the Leray-Hopf weak solution). To show this, he invented a revolutionary idea to construct a weak solution. Not only for constructing this solution, but he also investigated a limitation of this solution by introducing an epoch of regularity. While this paper was written 90 years ago, I can see how he thought about this problem, and his perspective amuses me. Modern PDE theory did not deviate that much from this paper.

I heard that someone created a review website for an AI paper from Stanford. If the user uploads the paper as a PDF, the AI will evaluate it and provide feedback. Link

It seems this tool was created because publishing a paper on AI is too competitive. I don’t know how to think. I believe that humankind chose to be destructive. Now we have lost lots of critical thinking, and many writings have become homogeneous.

90% of research papers will be forgotten. Many people will not listen to how to write a good paper in the extreme era of publish-or-perish.

Now I kept thinking how to write a good article to touch the reader. That’s why I kept reading literature these days to get some idea.

College Education in the advent of Generative AI

2025-11-27T21:00:40+00:00

The original article is written in Korean by myself Link. I put the translation generated by Claude. I polished some word choice which generative AI cannot translate properly.

I will try to rewrite this article in the future within December.

Since ChatGPT emerged three years ago, GenAI has made remarkable progress and has deeply influenced at least university education. I believe fundamental changes in the traditional education system are necessary, but there doesn’t seem to be much discussion about what direction we should take. In this essay, I want to organize my thoughts on the direction university-level education should pursue.

Let’s imagine writing a novel based on the emotion of fear. I tried it with Gemini. I entered the prompt: “I want to write a horror novel. What suggestions would you make for initial settings or a draft?” Gemini replied that the main character should be an ordinary person readers can empathize with, and when they face an abnormal horror situation, readers should also feel fear. Additionally, I should give the main character a fatal weakness or flaw. Finally, I must define the object of fear, and the object of fear (monster, ghost, curse, etc.), but I should not be described in detail.

The first answer AI produces from a prompt is, as expected, average and predictably structured. However, works that transcend their era—novels that people still read 200 years later—don’t fit the structure Gemini suggested. Around 1817, at Byron’s villa, while Percy Shelley and Claire Clairmont were trying to break the monotony of rainy days, the idea came up to write horror stories. Mary struggled to write a novel because of a lack of inspiration, but eventually wrote one after an interesting dream. It was called “Walking Dream,” now known as Frankenstein. Though written 200 years ago, it’s the first science fiction novel, and rather than being analyzed merely as fantasy, it’s a novel beloved to this day for raising many questions, adapted into plays and, recently, into films.

While we could speak of this novel’s originality in many ways, taking the theme of “horror” and extracting “the primal emotions of pain and loneliness when a new life form created by science and technology is abandoned at birth, struggles in that abandonment, tries to assimilate into society but isn’t accepted because of its appearance”. This is the discovery of subject matter that transcends average knowledge and thinking. I believe this became a great novel that continues to raise many questions today, thanks to Mary Shelley’s life journey.

Current mathematics education at universities has a structure of introducing definitions, proving them, advancing through examples, checking one’s knowledge, and moving forward by solving practice problems as assignments. Incidentally, this educational composition became established about 70 years ago; in the past, students read without practice problems, understanding between the lines of the text and creating their own examples. However, with the universalization of university education and the development of engineering, the need to solve concrete practical problems emerged, and I understand that this led to the establishment of current teaching methods. However, the birth of generative AI is shaking the foundation of this long-standing educational system. The practice-problem method of evaluating students turns out to be ineffective and it becomes problematic. In other words, the foundation of the educational system, knowledge verification and proficiency evaluation through homework, has been shaken.

So how should we solve this problem? We must now always assume AI usage. Therefore, I think attempting to block it entirely is an idea that goes against the times. While there are negative effects, education’s goal should be to maximize positive impacts and enable future generations to develop better ideas. GenAI is a new era’s search tool. Documents written in French or Russian are now more accessible, and when preparing classes, I can quickly research unfamiliar historical contexts, so I can deliver lectures that are more engaging for students. When writing papers and studying other papers, GenAI helps roughly grasp core ideas and kindly explains parts that are so basic I’m afraid to ask about, filling knowledge gaps. Despite these advantages, insisting only on past systems is no different from arguing that we should return to calculator-era exams because computers ruin students.

How was Mary Shelley able to write such a novel? Mary had intellectual soil that allowed access to much knowledge, studying literature and science by reading many books in her library under her open-minded father and mother, and she could write a novel inspired by Galvanism, which was popular in the late 18th century. With Galvani’s successful experiment of moving dead animal muscles using electricity, she could conceive the idea of creating life with electricity for Frankenstein’s birth. Additionally, she could write such a novel thanks to her husband, Percy Shelley’s, support. In 1800, women’s status was significantly lower than it is now, making it difficult to publish novels under one’s own name. Having someone who can support you gives great strength in taking on new challenges.

So how should university education proceed? We need to care more about students and stop viewing them merely as objects of evaluation. To do this, we need an era that pays more attention to students, especially those with a passion for learning, by focusing on how students can acquire new perspectives and understand more effectively. We must remind students who simply use AI to get answers and grades that they’re wasting their time. If they use knowledge not taught in class, they should be severely punished, but we shouldn’t discourage the attitude of trying to learn new knowledge by using AI to write their answers more perfectly.

However, as a practical matter, in the case of basic mathematics courses, it’s true that if these courses are shaken, there will be great difficulty in digesting other science and engineering classes. For this reason, for basic courses only, it may be necessary to change the system to periodic problem-solving assignments, build basic stamina through handwriting, and evaluate through quizzes. Additionally, during that process, AI supports repetitive learning, and instructors use the saved time to provide customized guidance to where students are stuck. This is the direction university education should head in the next 10 years. Now we need to focus on how we can teach those who could ask a daring question.

Since students in American universities are absent so often, I don’t know how to solve this. However, universities must now pay more attention not to the simple transmission of knowledge, but to the role of laying intellectual groundwork so that new imagination can unfold.

Blashcke-Santalo inequality and Aleksandrov estimates

2025-10-23T10:00:40+00:00

Yesterday, while I am attending ICERM conference on fluid dynamics and turbluence, I learned from Nestor that there is another way to look at the Aleksandrov maximum principle via convex geometry. I appreciate his passion for sharing this knowledge with me.

In my previous blog post, Aleksandrov estimate plays an important role in regularity theory for equations in nondivergence form or fully nonlinear equations. For simplicity, we consider a convex function $h:D\rightarrow\mathbb{R}$ defined on a convex body $D$ satisfying $h=0$ on $\partial D$. Furthermore, we assume that the center of mass of $D$ is the origin. A classical Aleksandrov estimate is that

\begin{equation} \Vert h \Vert_{L^\infty(D)} \leq C_d|D|^{1/d}|\nabla h(D)| \end{equation} for some constant $C_d>0$.

There is an interesting connection with Blashcke-Santalo inequality and this estimate. To state this, let us define the polar dual \begin{equation*} D^*=\{x\in\mathbb{R}^d : x\cdot y \leq 1\quad \text{for all } y \in D\}. \end{equation*} In 1939, Mahler proved that \begin{equation*} 4^n (n!)^{-2}\leq |D| |D^*|\leq 4^n. \end{equation*} Later, Santalo proved that the upper bound has a precise geometric bound \begin{equation*} |D| |D^*|\leq |B_d|^2, \end{equation*} where $B_d$ is the $d$-dimensional unit ball. The inequality becomes equality when $D=B_d$. After this work, several interesting improvements were made to determine the precise lower bound for $|D| |D^*|$. I do not have a clear picture of how researchers could obtain optimal estimates by using the geometry of Banach spaces (this is slightly unclear to me). Bourgan-Milman proved the following estimate, which is now called Blashcke-Santalo inequality.

ongoing….

Smoothing estimate for heat equation with compactly supported initial data

2025-01-09T21:00:40+00:00

Today, I learned from Sammer the following technique in the winter school held at Stony Brook University. For beginner, I decide to give a detailed proof as possible I can (although some argument can be omitted in a professional level). The motivation of this calculation can be found in the recent paper due to Bedrossian-He-Iyer-Wang.

Let us consider \begin{equation} \partial_t u- u_{xx}=0\quad \text{in } (0,\infty)\times \mathbb{R},\quad u=u_0\quad \text{on } \{t=0\}\times \mathbb{R}. \end{equation} It is well-known that if $u_0 \in L^2(\mathbb{R})$, then the solution $u$ is instantaneously smooth for $t>0$. However, the solution does not have good norm control in derivatives. However, if we assume that the initial data has a compact support, say $\mathrm{supp}\, u_0 \subset (-1/4,1/4)$, then we can show that $$ u\in L^{\infty}(0,\infty;L^{2}((-2,2)^c). $$

To show this, we first recall the global estimate for the heat equation: \begin{equation} \int_{\mathbb{R}}|u(t,x)|^2\,{dx}+2\int_0^t \int_{\mathbb{R}} |\partial_x u(t,x)|^2 \,dx=\int_{\mathbb{R}} |u_0(x)|^2 \,dx. \end{equation}

Choose a cut-off function $\chi$ so that $\chi=1$ in $(-1,1)$ and $\chi=0$ outside $(-2,2)$. For $h\neq 0$, define $u_h(t,x) = (u(t,x+h)-u(t,x))/h$. Then $u_h$ satisfies $$ \partial_t (u_h)-(u_h)_{xx}=0\quad \text{in } (0,\infty)\times \mathbb{R},\quad u_h=(u_0)_h\quad \text{on } \{t=0\}\times \mathbb{R}. $$

Choose $\chi$ so that $0\leq \chi\leq 1$, $\chi=0$ in $(-1,1)$, and $\chi=1$ on $(-2,2)^c$. Multiplying $u_h \chi$ and taking integration by part, we get \begin{equation} \begin{aligned} &\frac{1}{2}\int_{\mathbb{R}} |u_h(t)|^2 \chi \,dx + \int_0^t \int_{\mathbb{R}} |\partial_x (u_h)|^2 \chi \,dx ds \\ &=\frac{1}{2}\int_0^t \int_{\mathbb{R}} |u_h|^2 \chi'' \,dxds+\frac{1}{2}\int_{\mathbb{R}} |u_h(0)|^2 \chi\,dx. \end{aligned} \end{equation} For sufficiently small $h$, we note that $u_h(0,x)\chi(x)=0$ since $\mathrm{supp}\, u_0 \subset (-1,1)$ and $\chi=0$ on $(-1,1)$. Hence it follows that $$ \frac{1}{2}\int_{\mathbb{R}} |u_h(0)|^2 \chi\,dx=0.$$ Hence by a method of finite difference, we get $$ \int_{\mathbb{R}} |u_x|^2 \chi \,dx+\int_0^t \int_{\mathbb{R}} |\partial_{xx} u|^2 \chi \,dxds \leq C \int_0^t \int_{\mathbb{R}} |u_x|^2|\chi''| \,d{x}ds. $$ Moreover, it follows from the energy estimate that $$ \int_{\mathbb{R}} |u_x(t)|^2 \chi \,dx+\int_0^t \int_{\mathbb{R}} |\partial_{xx} u|^2 \chi \,dxds \leq C \int_{\mathbb{R}} |u_0|^2 \,d{x}. $$ Hence it follows that $u \in L^\infty((0,\infty);H^1((-2,2)^c))$. Then by induction, we can further show that $u\in L^\infty ((0,\infty);H^k((-3,3)^c))$ for any $k$.

In fact, we can obtain better smoothing estimate. It belongs to Gevrey $1/s$-class $G^s$ for $0

Note that $$ \partial_y^2 (\chi_{n+1}^2)\leq C (n+1)^{2(1+\sigma)}\chi_n^2.$$ So it follows that \begin{equation} \begin{aligned} &\int_{\mathbb{R}}|\partial_x^{n+1} u(t)|^2 \chi_{n+1}^2 d{x}+\int_0^t \int_{\mathbb{R}}|\partial_x^{n+2}u|^2 \chi_{n+1}^2 dxds\\ &\leq C(n+1)^{2(1+\sigma)}\int_0^t \int_{\mathbb{R}}|\partial_x^{n+1}u|^2 \chi_n^2 dxds. \end{aligned} \end{equation} Since $\sigma<1/s-1$, it follows that \begin{equation} \begin{aligned} &\left(\frac{1}{(n+1)!}\right)^{2/s}\int_{\mathbb{R}}|\partial_x^{n+1} u(t)|^2 \chi_{n+1}^2 d{x}+\left(\frac{1}{(n+1)!}\right)^{2/s}\int_0^t \int_{\mathbb{R}}|\partial_x^{n+2}u|^2 \chi_{n+1}^2 dxds\\ &\leq C\left(\frac{1}{n!}\right)^{2/s}\int_0^t \int_{\mathbb{R}}|\partial_x^{n+1}u|^2 \chi_n^2 dxds. \end{aligned} \end{equation}

If we set \begin{equation} \begin{aligned} a_n &=\int_{\mathbb{R}}|\partial_x^{n} u(t)|^2 \chi_{n}^2 d{x},\\ b_n &=\int_0^t\int_{\mathbb{R}}|\partial_x^{n+1} u|^2 \chi_{n+1}^2 d{x}ds, \end{aligned} \end{equation} then by induction, we have $$ b_n \leq C^n (n!)^{2(1+\sigma)}\Vert b_0\Vert_{L^2} $$ and so $$ \sum_{n=0}^\infty \left(\frac{1}{(n+1)!}\right)^{2/s}a_{n+1} \leq \Vert{b_0}\Vert_{L^2} \sum_{n=0}^\infty C^n \left(\frac{1}{n!} \right)^{2/s-2(1+\sigma)} .$$ Since $2/s-2(1+\sigma)>0$, the series converges. This implies that $$ \sum_{n=0}^\infty \left(\frac{1}{n!}\right)^{2/s}\int_{(-1/2,1/2)^c} |\partial_x^n u(t)|^2 dx<\infty.$$ Hence $u$ belongs to $L^\infty(0,\infty;G^{s}((-1/2,1/2)^c))$.

As we can see, the above argument does not work if $s=1$.

Lower bound for $L^2$ norm of second-order polynomials on a ball

2024-10-05T10:00:40+00:00

The following calculation seems quite elementary but it is a bit hard to find it anywhere. However, I need to do it because of new paper.

Set $$ \phi(r,X_0)=\sup_{t\in (t_0-r^2,t_0)} \frac{c}{r^2}\left(\frac{1}{|B_r|}\int_{B_r(x_0)} |u(t,x)-q_{X_0,r}(x)|^2 \,d{x}\right)^{1/2} $$ for some $q_{X_0,r}(x)=a_{X_0,r}+ b_{X_0,r} \cdot (x-x_0) + (x-x_0)^T C_{X_0,r}(x-x_0)$.

If we take

\begin{equation} \begin{aligned} q_{X_0,r}-q_{X_0,\kappa r} &= (a_{X_0,r}-a_{X_0,\kappa r}) + (b_{X_0,r}-b_{X_0,\kappa r})\cdot (x-x_0)\\ &\quad+ (x-x_0)^T (C_{X_0,r}-C_{X_0,\kappa r})(x-x_0), \end{aligned} \end{equation}

then a change of variable reduces the problem into

$$ \frac{c}{r^2} \left(\frac{1}{|B_r|}\int_{B_r(x_0)} |q(x)|^2 \,d{x}\right)^{1/2}=\frac{c}{r^{d/2+2}} \left(\int_{B_r}\left|a+b\cdot y + y^T Ay\right|^2 \,d{y}\right)^{1/2}. $$

Let

$$ q(x)=a+ b\cdot x + x^T A x. $$

Then we claim that

$$ \frac{c}{r^2}\left(\frac{1}{|B_r|}\int_{B_r} |q(x)|^2 \,d{x}\right)^{1/2}\geq c |A|.$$

Note that

\begin{equation} \begin{aligned} \left|a+b\cdot y + y^T Ay\right|^2&=a^2 + (b\cdot y)^2 +|y^T A y|^2 \\ &\quad+2 a (b\cdot y) + 2a(y^T Ay)+2(b\cdot y) y^T A y. \end{aligned} \end{equation} Using polar coordinates and oddness of $(b\cdot y)$ and $2(b\cdot y)y^T Ay$, we have \begin{equation} \begin{aligned} &\int_{B_r} |a + b\cdot y + y^T A y|^2 \,d{y}\\ &=\int_{B_r} a^2 +(b\cdot y)^2 +|y^T Ay|^2 + 2a(y^T Ay)\,d{y}\\ &=\int_{\mathbb{S}^{d-1}}\int_0^r \left(a^2 + \rho^2(b\cdot \omega)^2+\rho^4 |\omega^T A \omega|^2 +2a\rho^2(\omega^T A \omega)\right) \rho^{d-1}d\rho d\sigma(\omega)\\ &=\int_{\mathbb{S}^{d-1}} \frac{a^2}{d}r^d+\frac{(b\cdot \omega)^2}{d+2} r^{d+2}+\frac{|\omega^T A\omega|^2}{d+4}r^{d+4}+\frac{2a(\omega^T A\omega)}{d+2}r^{d+2} d\sigma(\omega). \end{aligned} \end{equation}

By Young’s inequality, we have

$$ |a(\omega^T A\omega)r^{d+2}| \leq \varepsilon a^2 r^d +\frac{1}{4\varepsilon} (\omega^T A\omega)^2r^{d+4}$$

for any $\varepsilon>0$. This implies that

\begin{equation} \begin{aligned} &\int_{B_r} |a + b\cdot y + y^T A y|^2 \,d{y}\\ &\geq \int_{\mathbb{S}^{d-1}} \left(\frac{1}{d}-\frac{2\varepsilon}{d+2}\right)a^2 r^d +\frac{(b\cdot \omega)^2}{d+2}r^{d+2}\,d{\sigma(\omega)}\\ &\quad+ \int_{\mathbb{S}^{d-1}} \left(\frac{1}{d+4}-\frac{1}{2\varepsilon(d+2)}\right)(\omega^T A\omega)^2 r^{d+4} d\sigma(\omega). \end{aligned} \end{equation}

We can choose $\varepsilon>0$ so that the numbers in parenthesis become positive numbers since $d(d+4)<(d+2)^2$. This implies that

\begin{equation} \begin{aligned} \fint_{B_r} |a+b\cdot y +y^T A y|^2 \,d{y}&\geq cr^4 \int_{\mathbb{S}^{d-1}} (\omega^T A\omega)^2 \,d{\sigma(\omega)}. \end{aligned} \end{equation}

On the other hand, note that

$$ \int_{\mathbb{S}^{d-1}} \omega_i \omega_j \omega_k \omega_l \,d{\sigma(\omega)}=\begin{cases} 1 &\quad \text{if } (i,j)=(k,l)\\ 0 &\quad \text{otherwise}. \end{cases} $$

Hence it follows that

$$ \int_{\mathbb{S}^{d-1}} (\omega^T A\omega)^2 \,d{\sigma(\omega)} =\int_{\mathbb{S}^{d-1}} (\omega_i A_{ij} \omega_j)(\omega_k A_{kl} \omega_l) \,d{\sigma(\omega)}\geq c\sum_{i,j} A_{ij}^2. $$

This implies the desired result.

Poincare inequality on weighted spaces

2023-07-04T10:00:40+00:00

In this note, we prove Poincar\'e-Sobolev inequality in weighted spaces. Such inequality was first observed by Fabes-Kenig-Serapioni and later simplified by Chiarenza-Frasca.

Theorem. Let $10$. Then there exists a constant $N=N(d,p,[w]_{A_p})>0$ such that $$ \left(\frac{1}{w(B_R)} \int_{B_R} |u|^p w dx \right)^{1/p}\leq NR \left(\frac{1}{w(B_R)} \int_{B_R} |\nabla u|^p w dx \right)^{1/p}$$ for all $u\in C_0^\infty(B_R)$. Here $A_p$ denotes the Muckenhoupt $p$-admissible weight.

In this note, we follow the proof of Chiarenza-Frasca. The idea follows from an idea of Hedberg. It suffices to show that $$ \left(\frac{1}{w(B_R)} \int_{B_R} |I u|^p w dx \right)^{1/p}\leq NR \left(\frac{1}{w(B_R)} \int_{B_R} |u|^p w dx \right)^{1/p},$$ where $I f$ denotes the Riesz potential of order $1$: $$ I f(x)=c \int_{\mathbb{R}^d} \frac{f(y)}{|x-y|^{d-1}}dy. $$ Decompose $$ If(x)=I_\varepsilon + II_{\varepsilon},$$ where $$ I_\varepsilon =c\int_{|x-y|<\varepsilon} \frac{1}{|x-y|^{d-1}} f(y)dy $$ and $$ II_\varepsilon =c\int_{|x-y|\geq \varepsilon} \frac{1}{|x-y|^{d-1}} f(y)dy. $$ On the one hand, we decompose the integrand into dyadic shell to estimate that \begin{align} |I_\varepsilon|&\leq \sum_{j=0}^\infty \int_{\varepsilon 2^{-j-1} \leq |x-y|\leq \varepsilon 2^{-j}} \frac{|f(y)|}{|x-y|^{d-1}}dy\\ &\approx \frac{\varepsilon 2^{-j}}{(\varepsilon 2^{-j})^d}\sum_{j=0}^\infty \int_{|x-y|\leq \varepsilon 2^{-j}} |f(y)| dy\\ &\leq N \varepsilon Mf(x) \sum_{j=0}^\infty 2^{-j}\\ &\leq N \varepsilon Mf(x). \end{align} On the other hand, it follows from Holder's inequality that \begin{align} II_\varepsilon&\leq \Vert f \Vert_{L_{p,w}(B_R)} \left(\int_{\{|x-y|>\varepsilon\}\cap B_R} |x-y|^{(1-d)p'} w^{-1/(p-1)}dy\right)^{1/p'}. \end{align} By reverse Holder property of $A_p$ weight, there exists $1p/q$ so that $w\in A_q$. Hence it follows that $$ II_\varepsilon (x)\leq c \Vert f \Vert_{L_{p,w}(B_R)}\left(\int_{B_R} w^{-1/(q-1)} dy\right)^{(q-1)/p} \varepsilon^{1-dq/p}. $$ Hence we get $$ If(x)\leq N \varepsilon M f(x) + N \Vert f \Vert_{L_{p,w}(B_R)} \left(\int_{B_R} w^{-1/(q-1)} dy\right)^{(q-1)/p} \varepsilon^{1-dq/p}. $$ Now we minimize with respect to $\varepsilon$ to get $$ If(x)\leq N[Mf(x)]^{1-p/nq} \Vert f \Vert_{L_{p,w}(B_R)}^{p/dq} \left(\int_{B_R} w^{-1/(q-1)} \right)^{(q-1)/dq}. $$ By taking $L_{p,w}(B_R)$-norm to get $$ \Vert I f \Vert_{L_{pk,w}(B_R)} \leq N \Vert f \Vert_{L_{p,w}(B_R)} \left(\int_{B_R} w^{-1/(q-1)} dy \right)^{(q-1)/dq}, $$ where $k=dq/(dq-p)$. Then we have \begin{align} &\left(\frac{1}{w(B_R)} \int_{B_R} |If|^{pk} w dx \right)^{1/{pk}}\\ & \leq N\left(\frac{1}{w(B_R)} \int_{B_R} |f|^p w dx \right)^{1/p} w(B_R)^{\frac{1}{p}\left(1-\frac{1}{k}\right)} \left(\int_{B_R} w^{-\frac{1}{q-1}} dx \right)^{(q-1)/dq}. \end{align} By $A_q$-condition, we have \begin{align} &w(B_R)^{\frac{1}{p}\left(1-\frac{1}{k}\right)} \left(\int_{B_R} w^{-\frac{1}{q-1}} dx \right)^{(q-1)/dq}\\ &=|B_R|^{\frac{1}{dq}}|B_R|^{\frac{q-1}{dq}}\left(\frac{1}{|B_R|} \int_{B_R} w dx \right)^{\frac{1}{dq}} \left( \frac{1}{|B_R|} \int_{B_R} w^{-\frac{1}{q-1}} dx \right)^{\frac{q-1}{dq}}\\ &\leq NR [w]_{A_q}^{1/(dq)}. \end{align} This implies that \begin{align} \left(\frac{1}{w(B_R)} \int_{B_R} |If|^{pk} w dx \right)^{1/{pk}}&\leq N R \left(\frac{1}{w(B_R)} \int_{B_R} |f|^{p} w dx \right)^{1/{p}}. \end{align} Then the desired result follows from Jensen's inequality.

A representation of functions $f$ in negative parabolic Sobolev spaces on bounded domain

2023-07-04T10:00:40+00:00

While I am preparing new paper, I found that I need to use the parabolic analogue result of the following proposition.

Proposition. Let $\Omega$ be a bounded domain in $\mathbb{R}^d$, $d\geq 2$ and $1=-\int_\Omega \mathbf{F}\cdot\nabla \varphi$$ for all $\varphi \in C_0^\infty(\Omega)$.

Proof of this fact requires Riesz representation theorem on $W^{-1}_p(\Omega)$ and Calderon-Zygmund estimates for Poisson equations, see Kim-Tsai [Lemma 3.9,KT20] A similar strategy also works for negative parabolic Sobolev spaces: recall that for $1< s,q<\infty$, we define $\mathbb{H}^{-1}_{s,q}((S,T)\times \Omega) = \{ u : u=f+\mathrm{div} F\quad\text{in } (S,T)\times \Omega \}.$

Proposition. Let $\Omega$ be a bounded domain in $\mathbb{R}^d$, $d\geq 2$ and $1
Proof. The idea is essentially the same with some modification. Since $\Omega$ is bounded, choose $R>0$ so that $\Omega \subset B_{2R}$. Fix a.e. $t\in (S,T)$. Then there exists a unique $v(t)\in\mathring{W}^1_q(B_{2R})\cap W^{2}_q(B_{2R})$ satisfying $$ -\Delta v(t) = f \quad \text{in } B_{2R}\quad v(t)=0\quad \text{on } \partial B_{2R}.$$ A bit problematic one is that it is not clear whether $(t,x)\mapsto v(t,x)$ is measurable. To handle this issue, we use a Green function of $-\Delta$. Let $G$ be a Green function of $-\Delta$ on $B_{2R}$. Then $$v(t,x)=\int_{B_{2R}} G(x,y) f(t,y)dy.$$ From this representation formula, it is easy to show that $v$ is measurable in $(t,x)$. Define $$ H(t,x) = \nabla_x v(t,x)|_{\Omega} + F(t,x).$$ Then it is easy to show that $$ \mathrm{div}\, H = f+\mathrm{div}\, F=u\quad\text{in } (S,T)\times \Omega$$ and $$\Vert H\Vert_{L_{s,q}((S,T)\times \Omega)}\leq N (\Vert{f}\Vert_{L_{s,q}((S,T)\times \Omega)}+\Vert{F}\Vert_{L_{s,q}((S,T)\times \Omega)}),$$ where the constant $N$ depends on the diameter of $\Omega$, $s$, $q$, and $d$. This completes the proof.

Green function of kinetic Kolmogorov-Fokker-Planck equations

2022-09-26T10:00:40+00:00

Here we consider a more simple Kolmogorov-Fokker-Planck equations \begin{equation} \partial_t u+ v\cdot D_x u -\Delta_v u =\mathrm{div}_v H+h. \end{equation} To find the fundamental solution of the equation, we consider the following Cauchy problem: \begin{equation} \left\{ \begin{alignedat}{2} \partial_t u+v\cdot D_x u-\Delta_v u&=h,\\ u(0)&=g. \end{alignedat} \right. \end{equation} By taking the Fourier transform in $(x,v)\mapsto (k,\xi)$, we have \begin{equation} \left\{ \begin{alignedat}{2} \partial_t \hat{u}-k\cdot D_{\xi} \hat{u}+|\xi|^2 \hat{u}&=\hat{h}(t,k,\xi),\\ \hat{u}(0)&=\hat{g}. \end{alignedat} \right. \end{equation} Along the characteristic $t\mapsto (t,k,\xi-tk)$, we have \begin{equation} \frac{\partial}{\partial t}\left(\hat{u}(t,k,\xi-tk \right))+|\xi-tk|^2 \hat{u}(t,k,\xi-tk)=\hat{h}(t,k,\xi-tk) \end{equation} Set \[ e(t,k,\xi)=\exp\left(-\int_0^t |\xi-\tau k|^2 d{\tau}\right).\] and note that \[ \frac{\partial}{\partial t} e(t,k,\xi)^{-1} =e(t,k,\xi)^{-1}|\xi-tk|^2. \] Multiplying $e(t,k,\xi)^{-1}$ with (4), we get \[ \frac{\partial}{\partial t}\left[e(t,k,\xi)^{-1}\hat{u}(t,k,\xi-tk) \right]=\hat{h}(t,k,\xi-tk)e(t,k,\xi)^{-1}. \] By taking integration with respect to $t$, we have \begin{equation} \begin{aligned} \hat{u}(t,k,\xi-tk)&=\hat{g}(k,\xi)e(t,k,\xi)\\ &\relphantom{=}+\int_0^t \hat{h}(s,k,\xi-sk)e(t,k,\xi)e(s,k,\xi)^{-1}ds. \end{aligned}\end{equation} Note that \[ e(t,k,\xi)e(s,k,\xi)^{-1} = \exp\left(-\int_s^t |\xi-\tau k|^2 d{\tau}\right).\] If we shift $\xi-tk$ into $\xi$, we get. \begin{equation} \begin{aligned} \hat{u}(t,k,\xi)&=\hat{g}(k,\xi+tk)e(t,k,\xi+tk)\\ &\quad +\int_0^t \hat{h}(s,k,\xi+(t-s)k)\exp\left(-\int_s^t |\xi+(t-\tau)k|^2d\tau\right)ds. \end{aligned}\end{equation} Note also that \[ e(t,k,\xi+tk)=\exp\left(-\int_0^t |\xi+(t-\tau)k|^2 d{\tau}\right)=e(t,-k,\xi)\] and \[ \exp\left(-\int_s^t |\xi+(t-\tau)k|^2d\tau \right) =\exp\left(-\int_0^{t-s} |\xi+\tau k|^2 d\tau\right)=e(t-s,-k,\xi).\] Hence we finally get \begin{equation} \hat{u}(t,k,\xi)=\hat{g}(k,\xi+tk)e(t,-k,\xi)+\int_0^t \hat{h}(s,k,\xi+(t-s)k)e(t-s,-k,\xi)ds. \end{equation} To take the inverse Fourier transform, we recall the following lemma.

Lemma. If $t,\beta>0$, then \[ \frac{1}{(2\pi)^{d/2}} \int_{\mathbb{R}^d} e^{-\beta|\xi|^2 t}e^{ix\cdot \xi} d{\xi}=\frac{1}{(2\beta t)^{d/2}}\exp\left(-\frac{|x|^2}{4\beta t}\right).\]

A change of variable shows that \begin{align} &\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{2d}} e(t,-k,\xi)e^{ik\cdot x} e^{i\xi \cdot v} d{k}d\xi\\ &=\frac{1}{(2\pi)^{d}t^d} \int_{\mathbb{R}^{2d}} \exp\left(-|\xi|^2 t-\frac{1}{12}|k|^2 t\right)e^{ix\cdot (k/t)} e^{iv\cdot (\xi-k/2)} d{k}d\xi \nonumber\\ &=\frac{1}{t^d} \left(\frac{1}{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp(-|\xi|^2 t)e^{i\xi \cdot v}d{\xi} \right)\left(\frac{1}{(2\pi)^{d/2}} \int_{\mathbb{R}^d} \exp\left(-\frac{1}{12}|k|^2 t\right)e^{i(x/t-v/2)\cdot k}d{k} \right) \nonumber\\ &=\frac{1}{t^d} \left(\frac{1}{(2t)^{d/2}} \exp\left(-\frac{|v|^2}{4t}\right)\right)\left(\frac{1}{(t/6)^{d/2}} \exp\left(-\frac{|x/t-v/2|^2}{t/3} \right) \right) \nonumber\\ &=\frac{3^{d/2}}{t^{2d}} \exp\left(-\frac{|v|^2}{4t}-\frac{3\left|x-\frac{t}{2}v\right|^2}{t^3}\right) \nonumber\\ &=\frac{3^{d/2}}{t^{2d}} \exp\left(-\frac{1}{4} \left|\frac{v}{t^{1/2}}\right|^2-{3\left|\frac{x}{t^{3/2}}-\frac{v}{2t^{1/2}}\right|^2}\right) \nonumber\\ &=\frac{3^{d/2}}{t^{2d}}\mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}}\right), \nonumber \end{align} where \[ \mathcal{G}(x,v)=\exp\left(-\frac{1}{4}|v|^2-3\left|x-\frac{v}{2}\right|^2\right).\] To proceed further, we define the modified convolution $*_t$ by \[ h*_t j(x,v)=\int_{\mathbb{R}^{2d}} h(y,w)j(x-y-tw,v-w)dwdy.\] Observe that if $\tilde{j}(x,v)=j(x+tv,v)$, then $h*_t j(x,v)=h*\tilde{j}(x-tv,v)$. Hence it follows from Young's convolution inequality that \begin{equation} \Vert{h*_t j}\Vert_{L^{r}(\mathbb{R}^{2d})}\leq \Vert{h}\Vert_{L^{p}(\mathbb{R}^{2d})}\Vert{j}\Vert_{L^{q}(\mathbb{R}^{2d})} \end{equation} independently of $t$, where $1+1/r=1/p+1/q$. If we take a Fourier transform to the convolution, then we have the following:

Proposition. For $t>0$, we have \[ \widehat{(h*_t j)}(k,\xi)=(2\pi)^d\hat{h}(k,\xi+tk)\hat{j}(k,\xi). \]

Proof. Recall that the Fourier transform of the convolution \[ (f*g)(x)=\int_{\mathbb{R}^d} f(x-y)g(y)d{y} \] is \[ \widehat{(f*g)}(\xi)=(2\pi)^{d/2}\hat{f}(\xi)\hat{g}(\xi).\] By taking Fourier transform, we have \begin{align} \widehat{h*_t j}(k,\xi)&=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^{2d}} (h*\tilde{j})(x-tv,v)e^{-ik\cdot x} e^{-i\xi \cdot v} d{x}dv\\ &=\frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} e^{-i\xi \cdot v} e^{-ik\cdot (tv)} \left[\int_{\mathbb{R}^d} (h*\tilde{j})(x,v)e^{-ik\cdot x}d{x} \right] dv \\ &=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^{2d}} e^{-i(\xi+tk)\cdot v} e^{-ik\cdot x} (h*\tilde{j})(x,v)d{x}dv\\ &=(2\pi)^{d} \hat{h}(k,\xi+tk)\hat{\tilde{j}}(k,\xi+tk). \end{align} Since $\tilde{j}(x,v)=j(x+tv,v)$, it follows that \begin{align} &=(2\pi)^d \hat{h}(k,\xi+tk)\hat{j}(k,\xi). \end{align} From this property and (7), we have \[ u(t,x,v)=(G*_t g)(x,v)+\int_0^t \int_{\mathbb{R}^{2d}} h(\tau)*_{t-\tau}G(t-\tau,\cdot,\cdot)d\tau, \] where \[ G(t,x,v)=\left(\frac{3}{4\pi^2 t^4} \right)^{d/2}\mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}} \right). \] We list some properties of the fundamental solution.

Proposition. The function $G$ and $\mathcal{G}$ have the following properties:

The function $\mathcal{G}$ is $C^\infty$ and decays polynomially at infinity. Moreover, $\mathcal{G}$ and all its derivatives are integrable in $\mathbb{R}^{2d}$.

For every $t>0$, $\int_{\mathbb{R}^{2d}} G(t,x,v)d{v}dx=1$.

Both functions are nonnegative: $G\geq 0$ and $\mathcal{G}\geq 0$.

For any $p\geq 1$, we have \begin{equation} \begin{aligned} \Vert{G(t,\cdot,\cdot)}\Vert_{L^{p}(\mathbb{R}^{2d})}&\approx t^{-d(1+1/2)(1-1/p)} \Vert{\mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}\\ \Vert{D_v G(t,\cdot,\cdot)}\Vert_{L^{p}(\mathbb{R}^{2d})}&\approx t^{-d(1+1/2)(1-1/p)-1/2} \Vert{D_v\mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}. \end{aligned}\end{equation}

Proof. (1) -- (3) are immediate from the definition. To show (4), a change of variable shows that \begin{equation} \begin{aligned} \Vert G(t,\cdot,\cdot)\Vert_{L^{p}(\mathbb{R}^{2d})}&=\left(\frac{3}{4\pi^2 t^4}\right)^{d/2} \left(\int_{\mathbb{R}^{2d}} \left|\mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}}\right)\right|^p d{x}dv\right)^{1/p}\\ &= \left(\frac{3}{4\pi^2}\right)^{d/2} t^{-2d(1-1/p)} \Vert{\mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}. \end{aligned}\end{equation} To show the second estimate, we note that \[ D_v G(t,x,v) = \left(\frac{3}{4\pi^2 t^4}\right)^{d/2} D_v \mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}}\right) \frac{1}{t^{1/2}}.\] A change of variable shows that \[ \Vert{D_v G(t,\cdot,\cdot)}\Vert_{L^{p}(\mathbb{R}^{2d})} = \left(\frac{3}{4\pi^2}\right)^{d/2}t^{-2d(1-1/p)-1/2}\Vert{D_v \mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}. \] This completes the proof.