Fix a "test" function $f(x)=x\exp(-x^2)$, which is nonzero except $x=0$. Suppose that $g$ is a function with some necessary regularity. Consider the convolution. $$ (f\ast g )(x)=\int_{-\infty}^{+\infty} f(y)g(x-y)dy. $$ Assume that $(f\ast g) (x)=0$ on an open interval $x\in (a,b)$. Does this imply that $g(x)$ takes a constant value almost everywhere on $\mathbb{R}$? (The converse is true due to the symmetry of $f(x)$.)
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Yes, because $f*g$ is indeed real-analytic (provided $g$ doesn't grow too fast at $\infty$), so that, being $0$ on an interval, it is $0$ everywhere. Then the Fourier transform of $f*g$, which is a Gaussian times the Fourier transform of $g'$, is $0$, and as the Gaussian is nonzero everywhere this implies $g'=0$ as a distribution.
-
$\begingroup$ Thanks for the reply. I just realized: would the Fourier transform of $f\ast g$ be the Fourier transform $\hat{f}$ of $f$ (which is derivative of Gaussian) times the Fourier transform of $g$. So the differentiation seems to be applied on $\hat{f}$ instead of $\hat{g}$. How does the argument work then? $\endgroup$– UchihaAug 14, 2015 at 11:24
-
$\begingroup$ $f$ itself is $\propto N'$, and $N'*g = N*(g')=(N*g)'$ $\endgroup$ Aug 14, 2015 at 15:19
-
$\begingroup$ Another question. In the argument it seems to require that $g$ is a tempered distribution. My feeling is that the statement might hold even for functions with exponential growth. Could you comment on this? Thank you. $\endgroup$– UchihaAug 17, 2015 at 14:39