# Altering L^{2} functions so that the FTC holds

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!

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I think I figured out the part about the measure of the set on which $f$ needs to be altered. Since $L^{2}$ convergence implies convergence in measure, the measure of the set where $f - f*\varphi_{\lambda} > \delta$ can be forced arbitrarily small by picking $\lambda$ large. – jmracek Apr 15 '12 at 21:19
Some ideas: the condition $\int_{\mathbb{R}} |\hat{f}(\xi)|^2 (1+|\xi|^2) d\xi <\infty$ implies that $\hat{f}$ is in $L^1$ by Cauchy-Schwartz. You can then show that Fourier inversion holds for $\hat{f}$ which gives that some version of $f \in C_0(\mathbb{R})$. The reason you need to modify by a set of measure zero is that there is no canonical member of an $L^2$ set of functions. They only exist up to sets of measure zero. For the remainder of the result, let $f_y$ be the translation of $f$ by $y$ and consider $\|\frac{1}{y}(f_y - f) - h\|_{L^2}$ and show $\exists h$ with this $\to 0$. – Chris Janjigian Apr 15 '12 at 21:32
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. – jmracek Apr 15 '12 at 22:26