$h(s)=\int_{-\infty}^{\infty}f(x)g(x-s) \, dx$ show that $h$ is uniformly continuous

Question

Let $f$ and $g$ be functions in $L^2(\mathbb{R})$. Define $h(s)=\int_{-\infty}^\infty f(x)g(x-s)\,dx$. Prove that $h(s)$ is uniformly continuous on $\mathbb{R}$.

I can't seem to make any meaningful progress on this one.

If I let $\epsilon > 0$, then I'm looking for a $\delta>0$ so that if $|s-t|<\delta$, we'll have

$$|h(s)-h(t)|=\left\vert \int_{-\infty}^\infty f(x)g(x-s)\,dx - \int_{-\infty}^\infty f(x)g(x-t) \, dx \right\vert = \left\vert \int_{-\infty}^\infty f(x)\left(g(x-s) - g(x-t)\right)\,dx \right\vert < \epsilon.$$

And that's about all I can think of for this one. I'd like to be able to say something about how close together $g(x-s)$ and $g(x-t)$ are, but all I know about them is that they're in $L^2$, so it's not obvious to me where to go from here.

I appreciate any thoughts anyone has.

Answer 1

For all $f\in L^p$ we have $\|f(x+k)-f(x)\|_p\to 0$ when $k\to 0$. If you have never seen this statement before, the proof goes like this: first show it for step functions, then use the fact that they are dense in $L^p$ to generalize it. This result is used a lot in problems about $L^p$ spaces.

Alright, so let $\varepsilon>0$. There is some $\delta>0$ such that $0<|k|<\delta$ implies $\|f(x+k)-f(x)\|_2<\varepsilon$. So now let $x,z\in\mathbb{R}$ which satisfy $|x-z|<\delta$. Then:

$$|h(x)-h(z)|=\left|\int_{\mathbb{R}}f(y)g(y-x) \, dy-\int_{\mathbb{R}} f(y)g(y-z) \, dy\right|=$$

$$=\left|\int_{\mathbb{R}}f(y)[g(y-x)-g(y-z)]\, dy\right|\leq $$

$$\leq \left(\int_{\mathbb{R}}|f(y)|^2 \, dy\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}}|g(y-x)-g(y-z)|^2\,dy\right)^{\frac{1}{2}}$$

Let $x-z=k$. Then we can continue:

$|h(x)-h(z)|\leq \|f\|_2 \left(\int_{\mathbb{R}}|g(y-z-k)-g(y-z) |^2 \, dy\right)^{\frac{1}{2}}$

And by changing variables $y-z=t$:

$$|h(x)-h(y)|\leq \|f\|_2 \left(\int_{\mathbb{R}}|g(t-k)-g(t)|^2 \, dy\right)^{\frac{1}{2}}$$

$$=\|f\|_2\times\|g(t-k)-g(t)\|_2\leq \|f\|_2\times\varepsilon$$

We used the assumption that $|k|=|x-z|<\delta$.

Answer 2

Here is a slightly general result that includes the OP with $p=2=q$.

Theorem: If $1/p +1/q=1$, $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$ and $g\in\mathcal{L}_q(\mathbb{R}^n,\lambda_n)$, then $f*g$ is uniformly continuous. If $1<p<\infty$ then $f*g\in\mathcal{C}_0(\mathbb{R}^n)$.

Here is a short proof:

We use $\operatorname{supp}(f)$ to denote the support of a function $f$, i.e. $\operatorname{supp}(f)=\overline{\{f\neq0\}}$.

Without lost of generality, we might assume that $1\leq p <\infty$. By Hölder's inequality and translation invariance of Lebesgue measure we have $$ \begin{align} |(f*g)(x+h)-(f*g)(x+k)|&\leq \int|(f(x+h-y)-f(x+k-y)||g(y)|\,dy\\ &\leq \|\tau_{-(k-h)}f-f\|_p\|g\|_q. \end{align} $$ Uniform continuity follows directly from the fact that the translation operator $\tau_h f\mapsto f(\cdot - h)$ satisfies $\lim\limits_{h\rightarrow}\|\tau_hf-f\|_p=0$ (This is well known; I present a short proof of this at the end of my answer to the OP).

To prove the last statement, let $\{f_k\}\cup\{g_k\}\subset\mathcal{C}_{00}(\mathbb{R}^n)$ such that $\lim\|f_k-f\|_p= 0=\lim_k\|g_k-g\|_q$ and $\operatorname{supp}(f_k)\cup\operatorname{supp}(g_k)\subset B(0;a_k)$. Then, $f_k*g_k\in\mathcal{C}_{00}(\mathbb{R}^n)$, $\operatorname{supp}(f_k*g_k)\subset B(0;2a_k)$ and, by H"older's inequality, $$ \|f*g-f_k*g_k\|_u\leq \|f-f_k\|_p\|g\|_q + \|f_k\|_p\|g-g_k\|_q. $$ We conclude that $f*g\in\mathcal{C}_0$ and hence, uniformly continuous.

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The following result is used on the proof of the answer to the OP.

Theorem: Suppose $1 \leq p < \infty$, and let $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$. Then, the mapping $\tau:\mathbb{R}^n\longrightarrow \mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, given by $t \mapsto \tau_t f=f(\cdot-t)$ is uniformly continuous.

Here is a short proof:

We first prove this lemma for continuous functions of compact support. Suppose that $g\in\mathcal{C}_{00}(\mathbb{R}^n)$ and that $\operatorname{supp}(g) \subset B(0,a)$ then, $g$ is uniformly continuous. Given $\varepsilon > 0$, by uniform continuity of there is a $0<\delta<a$ such that $|s-t|<\delta$ implies $$ |g(s) - g(t)| < (\lambda(B(0,3a)))^{-1/p}\varepsilon. $$ Hence, $$ \int |g(x-t) - g(x-s)|^p \, dx =\|\tau_t g - \tau_s g\|^p_p = \|\tau_{t-s}g -g\|^p_p < \varepsilon^p. $$ Therefore $t\mapsto \tau_tg$ is uniformly continuous. For general $f\in\mathcal{L}_p$, the conclusion follows from the density of ${\mathcal C}_{00}(\mathbb{R}^d)$ in $\mathcal{L}_p$.