$\textbf{Problem Statement} :$ Let $\{\phi_n\}$ be a sequence of continuous, real valued functions defined on $\mathbb{R}$ so that

$\phi_n(x)$ = 0 $\space \space \space \forall \vert x \vert \geq \frac{1}{n}$

$\phi_n(x) \geq 0 \space \space \space \forall x \in \mathbb{R}$

$\int_{-\infty}^{\infty} \phi_n(x) dx = 1$

For each continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$, define the sequence $f_n$ by

$f_n(x) = \int_{-\infty}^{\infty} \phi_n(x-y)f(y) dy$. $\space \space \space$ Show that $f_n \rightarrow f$ $\forall x \in \mathbb{R}$

So I made a slight change of variables and let $t = x-y$ so that $f_n(x) = \int_{\infty}^{\infty} \phi_n(t) f(x-t) dt$, and I notice $f(x) = f(x)(\int_{\infty}^{\infty} \phi_n(t) dt) = \int_{\infty}^{\infty} f(x)\phi_n(t) dt$.

So $\vert f_n(x) - f(x) \vert = \int_{\infty}^{\infty} (\vert f(x-t) - f(x) \vert)\phi_n(t) dt \leq \epsilon\int_{\infty}^{\infty} \phi_n(t) dt = \epsilon$.

However for one, I am not sure since $f$ is not uniformly continuous. In addition I don't need to show $f_n(x) \rightarrow f(x)$ uniformly in this part of the question. I'm to prove it how I se it pictorally, in that the $\phi_n$s (not necessarily tent\rectangle functions sorry for misnomer in title), converge to a dirac delta function to show pointwise convergence. Can I do that?

  • 1
    $\begingroup$ Hint: since $f()$ is continuous, given $x$ and an $\epsilon$ you can find a $\delta$-neighborhood of $x$ s.t. $\forall y\in(x-\delta,x+\delta), |f(y)-f(x)|\lt\epsilon$. Now take $n\gt1/\delta$ and bound $f_n(x)$ between $f(x)-\epsilon$ and $f(x)+\epsilon$... $\endgroup$ – Steven Stadnicki Apr 12 '18 at 19:07
  • $\begingroup$ (This is roughly the argument you have, but phrased more specifically.) $\endgroup$ – Steven Stadnicki Apr 12 '18 at 19:10

Using continuity of $f$, for given $\epsilon$ there is some $\delta$ such that $\forall |t|<\delta$, $|f(x-t) - f(x)| < \epsilon$. Now, take $N > \frac{1}{\delta}$, so $\forall n > N$,

$$\begin{align}\vert f_n(x) - f(x) \vert &= \int_{-\infty}^{\infty} \vert f(x-t) - f(x) \vert\phi_n(t) \, dt\\ &\leq \int_{-\frac{1}{n}}^{\frac{1}{n}}\vert f(x-t) - f(x) \vert \phi_n(t)\,dt \\ &\leq \epsilon \int_{-\frac{1}{n}}^{\frac{1}{n}} \phi_n(t)\,dt\\&=\epsilon\end{align}$$ as required.

The key point is that we know $\phi_n(x) = 0$ for all $|x| \geq \frac{1}{n}$, so we only need to look at $f$ in a small region, and hence only continuity is required instead of uniform continuity.


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