Proof that $x \mapsto \int |u(x+y)-u(y)|^p \lambda^n(dy)$ is continuous

Question

I am looking at the proof of the following theorem, but I have a question about the last part.

Let $u \in \mathscr{L}^p(\lambda^n), p \in [1,\infty)$.

The map $x \mapsto \int |u(x+y)-u(y)|^p \lambda^n(dy)$ is continuous.

Proof. Let $phi \in C_c(\mathbb{R}^n)$ then we can easily show using dominated convergence that $$x \mapsto I(\phi ; x):= \int |\phi(x+y)-\phi(y)|^p \lambda^n(dy) = \Vert \phi(x+\cdot)-\phi(\cdot)\Vert_p^p$$ is continuous. Now let $u \in \mathscr{L}^p(\lambda^n)$. Using the density of $C_c(\mathbb{R}^n)$, we find a sequence of $(\phi_n) \subset C_c(\mathbb{R}^n)$ such that $\lim_{n\to \infty} \Vert u-\phi_n\Vert_p=0.$ Since we have for any $x\in \mathbb{R}^n$,

$\Vert (\phi_n(x+\cdot)-\phi_n(\cdot))-(u(x+\cdot)-u(\cdot))\Vert_p \le \Vert \phi_n(x+\cdot)-u(x+\cdot)\Vert_p + \Vert \phi_n-u\Vert_p = 2\Vert \phi_n-u\Vert_p$, we see that $I(\phi_n;x)$ converges to $I(u;x)$ uniformly. In particular, $I(u;x)$ inherits the continuity of the $I(\phi_n;x)$.

My problem with this proof is, how do we even get uniform convergence from the inequality above? The above inequality basically gives the $L^p$ convergence of $I(\phi_n;x)$ to $I(u;x)$, which is not uniform convergence. So how do we complete the proof?

Answer 1

First, I wanted to note that unlike what you say the inequality does not show $L^p$ convergence of $I(\phi_n;x)$ to $I(u;x)$. These functions already involve integration of the respective functions $\phi$ and $u$, so their $L^p$ convergence would mean taking another integral on top of that, not something the inequality implies.

However, you are right that the uniform convergence is not implied directly or rather explicitly from the inequality. What you need is to derive it based on it. Namely,

$$ |I(\phi_n;x)^{1/p}-I(u;x)^{1/p}| := |\|\phi_n(x+\cdot)-\phi_n(\cdot)\|_p-\|u(x+\cdot)-u(\cdot)\|_p|. $$

We take here $I^{1/p}$ because otherwise in the RHS we would have the difference of the powers of norms, for which the reverse triangle inequality does not hold. For example, take $p=2,g=\tfrac{1}{2}f(x)$,

$$ |\|f\|_p^p-\|g\|_p^p|=\tfrac{3}{4}\|f\|_2^2>\|f-g\|_p^p=\tfrac{1}{4}\|f\|_2^2. $$

However, the reverse triangle inequality does hold for the norms themselves, so we further have

$$ |\|\phi_n(x+\cdot)-\phi_n(\cdot)\|_p-\|u(x+\cdot)-u(\cdot)\|_p| \le \|\phi_n(x+\cdot)-\phi_n(\cdot)-u(x+\cdot)+u(\cdot)\|_p. $$

And the inequality stated in the proof implies that

$$ I(\phi_n;x)^{1/p} \rightrightarrows I(u;x)^{1/p}. $$

From here you can already conclude that $I(u;x)$ is continuous.

Answer 2

Fix an $x'$. \begin{align*} |I(u,x)^{1/p}-I(u,x')^{1/p}|&\leq|I(u,x)^{1/p}-I(\phi_{n},x)^{1/p}|+|I(\phi_{n},x)^{1/p}-I(\phi_{n},x')^{1/p}|+|I(\phi_{n},x')^{1/p}-I(u,x')^{1/p}|, \end{align*} note that $|I(\phi_{n},x)^{1/p}-I(\phi_{n},x')^{1/p}|$ can be controlled by arbitrary small by the first result. Now \begin{align*} |I(u,x)^{1/p}-I(\phi_{n},x)^{1/p}|&=\bigg|\|u(x+\cdot)-u(\cdot)\|_{p}-\|\phi_{n}(x+\cdot)-\phi_{n}(\cdot)\|_{p}\bigg|\\ &\leq\|\phi_{n}(x+\cdot)-\phi_{n}(\cdot)-(u(x+\cdot)-u(\cdot))\|_{p}, \end{align*} so at the end $|I(u,x)^{1/p}-I(u,x')^{1/p}|$ can be controlled by arbitrary small. So $I(u,\cdot)^{1/p}$ is continuous at $x'$, then so is $I(u,\cdot)$ by taking exponent $p$.