doesn't seem correct

TIMsetsFIRE · **Joined:** Mon, 27 Apr 2009 15:11:34 UTC **Posts:** 156

I'm studying for analysis and I have the following question (given by a professor), which does not seem correct the way it is stated

Let $\phi(x): \mathbb{R} \to \mathbb{R}$ be a function, so that
$\int_{-\infty}^\infty \phi(x) dx = 1$ . Show that for any continuous function $f: \mathbb{R} \to \mathbb{R}$ we have that the sequence
$g_n(x) = n \int_{-\infty}^\infty \phi(n(x-y))f(y)dx \rightrightarrows f(x)$ on the compact sets of IR.

First, the way it is written doesn't even seem to be make sense because it is integrating with respect to x so I believe that the integral should actually be with resepect to y. But that doesn't seem to be true, for instance if I define
$\phi(x) = e^{-x}, x \ge 0$ and 0 otherwise. This integrates to 1. Moreoever, If I pick a function like f(x)= x^2 which is continuous on R, the function g_n doesn't even converge.
For instance,
$n \int_{0}^\infty \phi(n(x-y))f(y)dy = n \int_{0}^\infty e^{-n(x-y)}y^2dy = n e^{-nx}\int_{0}^\infty e^{ny}y^2dy$

I think the only way this problem could make sense is if the function $\phi$ is defined to have compact support.

outermeasure · **Posted:** Thu, 2 Jun 2011 06:29:11 UTC

TIMsetsFIRE wrote:

I'm studying for analysis and I have the following question (given by a professor), which does not seem correct the way it is stated

Let $\phi(x): \mathbb{R} \to \mathbb{R}$ be a function, so that
$\int_{-\infty}^\infty \phi(x) dx = 1$ . Show that for any continuous function $f: \mathbb{R} \to \mathbb{R}$ we have that the sequence
$g_n(x) = n \int_{-\infty}^\infty \phi(n(x-y))f(y)dx \rightrightarrows f(x)$ on the compact sets of IR.

First, the way it is written doesn't even seem to be make sense because it is integrating with respect to x so I believe that the integral should actually be with resepect to y. But that doesn't seem to be true, for instance if I define
$\phi(x) = e^{-x}, x \ge 0$ and 0 otherwise. This integrates to 1. Moreoever, If I pick a function like f(x)= x^2 which is continuous on R, the function g_n doesn't even converge.
For instance,
$n \int_{0}^\infty \phi(n(x-y))f(y)dy = n \int_{0}^\infty e^{-n(x-y)}y^2dy = n e^{-nx}\int_{0}^\infty e^{ny}y^2dy$

I think the only way this problem could make sense is if the function $\phi$ is defined to have compact support.

No! The question is essentially correct (and is how you construct approximations to Dirac delta).

For fixed $x\in\mathbb{R}$ and your
$\phi(x)=\begin{cases} e^{-x} & (x\geq 0)\\ 0 & (x<0) \end{cases},$
the function $y\mapsto\phi(n(x-y))$ is supported on $(-\infty,x]$ , so the first equality in the last equation should be
$\displaystyle g_n(x):=n \int_{-\infty}^\infty \phi(n(x-y))f(y)\,\mathrm{d}y = n \int_{-\infty}^\infty e^{-n(x-y)}1_{x>y}y^2\,\mathrm{d}y$
which then gives
$\displaystyle g_n(x)=ne^{nx}\int_{-\infty}^x e^{ny}y^2\,\mathrm{d}y=ne^{nx} \frac{e^{-nx}(nx(nx-2)+2)}{n^3} \to x^2$
pointwise, and uniformly on compacts.

The only typo in the question is integrate with respect to y, not x.

Edit: corrected $-\infty$ shouldn't be in the support. Also, spoke too soon about the question being OK. I think there will be a problem if f grows way too fast, e.g. $\phi(x)=e^{-\lvert x\rvert}/2$ , $f(x)=e^{x^2}$ , which means the convolution $\phi_n\ast f$ blows up dramatically for each fixed n. You want something like $f\in C_0(\mathbb{R})$ or $f\in C(\mathbb{R})\cap L^1(\phi)$ rather than just $f\in C(\mathbb{R})$ . Of course, it is easiest if $\phi$ has compact support....

TIMsetsFIRE · **Joined:** Mon, 27 Apr 2009 15:11:34 UTC **Posts:** 156

outermeasure wrote:

the function $y\mapsto\phi(n(x-y))$ is supported on $[-\infty,x]$ , .

Of course!!! I completely missed this when I was looking at it. Thank you!

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doesn't seem correct

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