Suppose $f \in L^2(0,2\pi)$ is such that there exists $v \in L^2(\mathbb{R})$ with
$$\int_{\mathbb{R}} f \phi' = \int_{\mathbb{R}} v \hat{\phi}$$ for all $\phi$ in Scwharz space on $\mathbb{R}$, and $\hat{\phi}$ is the fourier transform. Using the density of Schwarz in $L^2(\mathbb{R})$, show that there exists a bounded continuous function $g$ with $f=g$ a.e.
I'm a bit unclear as to how to begin this problem. The condition seems odd to me, I've been trying to somehow use an integration by parts, or the fact that this is a statements about inner products on $L^2$, but i'm unsure how to begin.
I have not worked out the entire problem, but I believe it should go something like this. You should use the fact that $$\int_{\mathbb{R}}v\hat{\phi}=\int_{\mathbb{R}}\hat{v}\phi.$$ This can be proven by writing the Fourier transform out as $\hat{v}(x)=\int_{\mathbb{R}}e^{-iyx}v(y)dy$ and switching the integrals (use Fubini). Next, perform an integration by parts. You will also need the fact that antiderivatives are continuous. Doing this should get you a continuous function $g$ such that $\langle f,\phi'\rangle=\langle g,\phi'\rangle$ for each $\phi$ in the Schwarz space. Using density then should get you that $f=g$ a.e.
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