If $\phi:\mathbb R\rightarrow\mathbb R$ with, $\int_{\mathbb R}\phi(x)\ dx=1$ and $\phi_{\delta}$ is defined by, $\phi_{\delta}(x)=\frac{\phi(\frac{x}{\delta})}{\delta}$. Prove that, for every continuous function $f:\mathbb R\rightarrow\mathbb R$ with compact support,$$f*\phi_{\delta}(x)\overset{(\delta\rightarrow 0)}\longrightarrow f(x)\ ,\forall x\in\mathbb R$$ (where * is the convolution). The rate of convergence may depend on $f$ and $x$.

$\phi_{\delta}$ looks like an approximation of unity, but it is not mentioned that it has a compact support.

Since $f$ has a compact support, $\exists k>0$, s.t. $f(x)=0$, if $\lvert x\rvert>k$. Then,

$f*\phi_{\delta}(x)=\int_{\lvert y\rvert\le k}\phi_{\delta}(x-y)f(y)dy+\int_{\lvert y\rvert>k}\phi_{\delta}(x-y)f(y)dy$

the second summand is zero, so consider the first;

$\int_{\lvert y\rvert\le k}\phi_{\delta}(x-y)f(y)dy\overset{\frac{x-y}{\delta}=u}=\frac{1}{\delta}\delta\int_{u\ge \frac{x-k}{\delta}}\phi(u)f(x-\delta u)du$

$\textbf{-can somebody correct the set of the integral, over which we integrate}$ because $u\ge \frac{x-k}{\delta}≠\mathbb R$ and i guess it must be the whole space.

Can we better choose here if $\lvert y\rvert\le k$ then $x-y\le x+k$ and $u=\frac{x-y}{\delta}\le \frac{x+k}{\delta}$ which implies that $(u\le \frac{x+k}{\delta})\rightarrow\mathbb R$ as $\delta\rightarrow 0$. Can you confirm this ?

$\textbf{-but where do we use the continuity}$ and does $f(x-\delta u)$ make sense ? (already answered)


Since $f$ is continuous you know that $f(x-\delta u)\to f(x)$ as $\delta\to 0$ for all $u\in \mathbb R$. Moreover $f$ is bounded hence $\phi(u)f(x-\delta u)$ can be dominated by $M\phi(u)$. So by Lebesgue theorem you conclude that $$ \int \phi(u) f(x-\delta u)\, du \to \int \phi(u) f(x)\, du = f(x). $$


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.