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.