# Show that $g(x) = \int_E f(x − t)d\lambda(t)$ is continuous on $\mathbb{R}$

Let f be a function in $$L^1 = L^1(\mathbb{R}, \lambda)$$, where $$\lambda$$ is the Lebesgue measure on $$\mathbb{R}$$. Let $$E$$ be a measurable set with $$\lambda(E) < \infty$$.

Prove that the following function is continuous on $$\mathbb{R}$$ $$g(x) = \int_E f(x − t)d\lambda(t)$$ You may use without proof that "for each $$\epsilon > 0$$ there exists a continuous function $$\phi$$ on R with compact support such that $$\|f − \phi\|_1 <\epsilon$$ ".

Is the following proof working?

fix $$\delta>0$$, and let the compact set to be such that $$A = \{x \in E : |x-y| \leq \delta ,\forall y \in E\}$$ so $$\lambda(A)\leq \delta$$. I have to show that $$\|g(x)-g(y)\|<\epsilon$$.

By above lemma there exists continuous function $$\phi$$ such that $$\|f(x-t) − \phi(x)\|_{L^1(A)} <\frac{\epsilon}{3}$$ and $$\|f(y-t) − \phi(y)\|_{L^1(A)} <\frac{\epsilon}{3}$$ , also let by continuity of $$\phi$$ let $$\|\phi(x)-\phi (y)\|_{L^1(A)}\leq \frac{\epsilon}{3}$$

\begin{align} \|g(x)-g(y)\|_{L^1(A)} & = |\int_Af(x-t)-\phi(x)+\phi(x)-\phi(y)+\phi(y)-f(y-t)d\lambda(t)|\\ & \leq \int_A |f(x-t)-\phi(x)|d\lambda(t)+\int_A |\phi(x)-\phi(y)|d\lambda(t)+\int_A |\phi(y)-f(y-t)|d\lambda(t)\\ & = \|f(x-t) − \phi(x)\|_{L^1(A)} + \|\phi(x) − \phi(y)\|_{L^1(A)} +\|f(y-t) − \phi(y)\|_{L^1(A)} \\ & \leq \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3}\\ & = \epsilon \end{align}

## 1 Answer

There is a considerable amount of confusion going on here. The fact you are given says that given $$\epsilon > 0$$, there exists a compactly supported continuous function $$\phi$$ such that $$\int_{\mathbb{R}} \vert f(t) - \phi(t) \vert d\lambda(t) < \epsilon.$$ Notice that in your integrals in the last few lines, $$\phi(x)$$ does not depend on the variable $$t$$ with respect to which you are integrating, so the estimates are not valid.

Secondly, you have chosen some set $$A$$ to integrate over. This doesn't make sense. You are trying to estimate $$\vert g(x) - g(y) \vert$$, where $$g$$ is a function taking a parameter $$x$$ to an integral over $$E$$ of a function depending on $$x$$. You are looking at the $$L^1$$ norm on $$A$$ of $$g(x) - g(y)$$, which is not the same as changing domain of integration in the integral defining $$g$$ to $$E$$.

There is also a major problem right at the start of your proof: for each $$x \in \mathbb{R}$$ you have a function $$t \to f(x - t)$$. There is no guarantee that all of these functions can be simultaneously approximated in $$L^1$$ by a single continuous, compactly supported $$\phi$$. What we can do is approximate the single function $$f$$, i.e., we can choose a continuous function $$\phi$$ with compact support $$\phi$$ such that $$\Vert f - \phi \Vert_{L^1(\mathbb{R})} < \frac{\epsilon}{3}$$.

For the sake of brevity let $$g_h(x) = \int_Eh(x - t)d\lambda(t)$$ for a function $$h \in L^1(\mathbb{R})$$. In particular $$g = g_f$$. Then for $$x,y \in \mathbb{R}$$ we have \begin{equation*} \begin{aligned} g(x) - g(y) &= g_f(x) - g_{\phi}(x) + g_{\phi}(x) - g_{\phi}(y) + g_{\phi}(y) - g_f(y), \end{aligned} \end{equation*} thus $$\vert g(x) - g(y) \vert \leq \vert g_f(x) - g_{\phi}(x) \vert + \vert g_{\phi}(x) - g_{\phi}(y) \vert + \vert g_{\phi}(y) - g_f(y) \vert,$$ so the goal is to show that the three terms above can be made $$< \frac{\epsilon}{3}$$ by choosing $$x,y$$ close enough.

The first term is $$\left\vert \int_E (f(x - t) - \phi(x - t)) d\lambda(t)\right\vert.$$ This is bounded by $$\Vert f - \phi \Vert_{L^1(\mathbb{R})}$$, which is $$< \frac{\epsilon}{3}$$ by our choice of $$\phi$$.

The third term is essentially identical. The middle term is $$\left\vert \int_E (\phi(x - t) - \phi(y - t)) d\lambda(t) \right\vert.$$ Here you need to use the following fact: since $$\phi$$ is continuous on $$\mathbb{R}$$ and compactly supported, in fact $$\phi$$ is uniformly continuous on $$\mathbb{R}$$. Since $$E$$ has finite measure, you can use uniform continuity to choose $$\delta > 0$$ such that if $$\vert x - y \vert < \delta$$, then the above is $$< \frac{\epsilon}{3}$$. Note that we don't have to fix any particular $$x$$ or $$y$$ here before choosing $$\delta$$ ($$g$$ is actually uniformly continuous).

The point of choosing a continuous compactly supported function really is to get a uniformly continuous, integrable function. But then we have the $$x$$ and $$y$$ to deal with as well as the $$f$$ and $$\phi$$. What we do is "change one at at a time" (first change $$x$$ to $$y$$ and then change $$f$$ to $$\phi$$ using the triangle inequality, leaving us with three quantities all of which we can estimate). This is a common technique throughout analysis: you want to estimate how some function $$A(x)$$ changes when the parameter $$x$$ is changed slightly. It would be much easier if $$A$$ had some nice properties, but it turns out $$A$$ can be approximated by some other function $$B(x)$$ which does have these nice properties. Then we use the triangle inequality: $$\vert A(x) - A(y) \vert \leq \vert A(x) - B(x) \vert + \vert B(x) - B(y) \vert + \vert B(y) - A(y) \vert.$$

• thanks, it was clear answer. Can you give like a graphical elaboration on the lemma? and on this part "There is no guarantee that all of these functions can be simultaneously approximated in $L^1$ by a single continuous, compactly supported $\phi$. What we can do is approximate the single function $f$, i.e., we can choose a continuous function $\phi$ with compact support $\phi$ such that $\|f−\phi\|_{L^1(R)}<\frac{\epsilon}{3}$." Oct 29, 2019 at 15:23