Let $f$ be an even continuous function with compact support such that
$$
\int f(t)\,\mathrm{d}t=1,
$$
and let $g$ be a bounded continuous function such that the convolution $f\star g$ satisfies the following equality
$$
(f\star g)(x)=g(x).
$$
How to prove that if $g$ has global minimum then $g$ is constant?
I thought about using the Convolution Theorem, but it seems it doesn't work in this case. Maybe, without the requirement that $g$ has global minimum, there is a way to prove that $g$ is linear?
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ If the global minimum for $g$ occurs on a set $E$, then write out $f\star g(x)$ for $x\in E$. $\endgroup$– Anthony QuasCommented Apr 24, 2020 at 18:53
-
$\begingroup$ @Anthony Quas, could you add some details? What should I get if I write f*g(x)? $\endgroup$– Anton SorokovskiyCommented Apr 24, 2020 at 18:59
-
$\begingroup$ @AntonSorokovskiy, you'll get something greater than $g(x) = \min g$, unless $E = \mathbb R$ (where I assume all this lives). This is not research level. $\endgroup$– LSpiceCommented Apr 24, 2020 at 19:09
-
$\begingroup$ @LSpice, I thought about it, but what if g can be negative on part of its support? $\endgroup$– Anton SorokovskiyCommented Apr 24, 2020 at 19:28
-
1$\begingroup$ Oh, I was, and perhaps @AnthonyQuas was also, assuming that $f$ was non-negative. $\endgroup$– LSpiceCommented Apr 24, 2020 at 22:23
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
2
The sought-after statement is wrong: $g$ can be non-constant. Fourier transforming your conditions we see that the Fourier transform $\hat{g}$ is supported at points where $\hat{f}$ is equal to $1$. We also see $\hat{f}(0)=1$ and $\hat{f}(-\xi)=\hat{f}(\xi)$. This motivates the following condition.
It should be trivial to construct an even function $f$ with compact support and such that $\int f(x)dx=1$ and $\int f(x)\cos(x)\,dx=1$. Then take $\hat{g}$ to be supported at $0$ and $\pm 2\pi$, for instance $g(x)=\cos(x)$. Then $$ (f\star g)(y) = \int f(x) g(y-x) dx = \cos(y) \int f(x) \cos(x)\,dx + \sin(y) \int f(x)\sin(x)\,dx = \cos(y) = g(y) $$ where I used $\int f(x)\cos(x)\,dx = 1$ by construction of $f$, and $\int f(x)\sin(x)\,dx = 0$ because $f$ is even.
-
3$\begingroup$ Per Fourier inversion, we could take for $f$ the extension by $0$ of the function $x \mapsto \frac1{2\pi}(1 + 2\cos(x) + \cos(2x))$ on $[-\pi, \pi]$. $\endgroup$– LSpiceCommented Apr 24, 2020 at 22:39
-
$\begingroup$ In order to use the machinery of the Fourier transform and deduce that $\hat f \hat g = \hat g$ from $f \star g = g$, you tacitly assume that $g$ is a function that admits a Fourier transform which is itself a function. But this excludes, among others, precisely the constant functions (the Fourier transforms of which are Dirac distributions). $\endgroup$– Alex M.Commented May 26, 2021 at 20:12