2
$\begingroup$

I'm studying for a real analysis exam by doing a previous year's test. I have come across one question which I have banged my head against for a few days but can't seem to make any progress.

I have two functions $f, g \in L^{2}(\mathbb{R})$ for which $\mathscr{F}(g)(\xi) = i\xi\hat{f}(\xi)$. The function $g(x)$ kind of wants to be the derivative of $f(x)$, but it isn't. We're talking about general $L^{2}$ functions for which the derivative doesn't have to exist. The function $f(x)$ also satisfies the requirement that \int_{\mathbb{R}} \vert \hat{f}(\xi)\vert^{2}(1+\vert\xi\vert^{2})\,d\xi < \infty.

I would like to show that one can alter $f(x)$ on a set of measure zero so that $f(x) - f(0) = \int_{0}^{x}g(t)\,dt$ for every $x \in \mathbb{R}$. My attempt so far has been to approximate $g(x)$ and $f(x)$ using a mollifier, show that the fundamental theorem of calculus holds for the approximations, then take a limit.

I'm stuck with the limiting process. I can't find a way to guarantee convergence $g*\varphi_{\lambda}(x)\rightarrow g(x)$ will be fast enough to exchange any limits and integrals. Is there some way I can guarantee $g(x)$ is continuous or uniformly continuous that I'm just not seeing? One technical detail that has me completely stumped is how to prove that any such alteration to $f(x)$ happens on a set of measure zero.

Any hints are much appreciated!

  • 0
    Thanks Chris! I think this was exactly what I needed. If $\hat{f}$ is integrable then I think that means that $f$ is uniformly continuous. In this case $f*\varphi_{\lambda}$ converges uniformly to $f$ as $\lambda \rightarrow \infty$. This gives me the proper convergence properties to make the rest of what I have worked out make sense.2012-04-15

0 Answers 0