Let $u$ be a Schwarz function in $\mathbb{R}^n$. Let $E$ be a distribution with support in the open unit ball, which satisfies $$ |\hat{E}(\xi)| \leq C (1 + |\xi|)^{-n}. $$ I am trying to prove the following inequality. For any $x \in \mathbb{R}^n$, $$ |(u * E) (x)|^2 \leq C \int_{|x - y| < 1} | u(y) |^2 dy, $$ where $C$ is a positive constant independent of $u$.
I am completely stuck on this problem... Any explanation is appreciated! Thank you!
Notice that $\hat{E}\in L^2$ since (for example) using polar coordinates $r=|\xi|$ we get
$$\int |\hat{E}(\xi)|^2d\xi \lesssim \int_0^\infty \frac{r^{n-1}}{(1+r)^{2n}}dr \leq 1+\int_1^\infty \frac{dr}{r^{n+1}}$$ Which converges since $n\geq1$.
Since $\hat{E}\in L^2$, the fact that the Fourier transform is an isomorphism of $L^2$ means that $E\in L^2$. By the support assumption we may write $E=E\chi_B$, where $B$ is the unit ball, then by Cauchy-Schwarz,
\begin{align*}|(u*E)(x)| &= \left|\int u(y)E(x-y)\chi_{B}(x-y) dy\right|\\ &= \left|\int (u\chi_{x-B})(y)E(x-y) dy\right| \\ &\leq \|E\|_{L^2}\left(\int |u\chi_{x-B}|^2 dy\right)^{1/2} \\ &= \|E\|_{L^2} \left(\int_{|x-y|<1}u(y)^2dy\right)^2 \end{align*}
Where in the second equality we used the fact that $x-y\in B \iff y\in x-B$, so $\chi_B(x-y) = \chi_{x-B}(y)$.
Edit. Let me provide more explanation to the claim that $E\in L^2$. The Fourier transform is an isomorphism of $L^2$, so in particular the inverse Fourier transform is well-defined and bounded as an operator $\mathcal{F}^{-1} : L^2\to L^2$, so $\mathcal{F}^{-1}\hat{E}\in L^2$. It then remains to show that $\mathcal{F}^{-1}\hat{E}$ and $E$ have the same action on Schwartz functions, which proves they're identical elements of the vector space of tempered distributions. If $\langle F,g \rangle = F(g)$ denotes the distribution evaluation pairing, Plancherel yields $\langle\mathcal{F}^{-1}\hat{E}, \phi\rangle = \langle\hat{E},\check{\phi}\rangle$, which equals $\langle E,\hat{\check{\phi}}\rangle = \langle E,\phi\rangle$ by definition. Alternatively, one could simply note that $\mathcal{F}:\mathcal{S}'\to\mathcal{S}'$ is bijective, so $\mathcal{F}^{-1}\mathcal{F}(E) = E$ as tempered distributions.