3
$\begingroup$

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}

$\endgroup$

1 Answer 1

2
$\begingroup$

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: $$\vert A(x) - A(y) \vert \leq \vert A(x) - B(x) \vert + \vert B(x) - B(y) \vert + \vert B(y) - A(y) \vert.$$

$\endgroup$
1
  • $\begingroup$ 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 $\|f−\phi\|_{L^1(R)}<\frac{\epsilon}{3}$." $\endgroup$
    – domath
    Oct 29, 2019 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.