An $f\in H^{1/2}$ with self-convolution, showing it is an $C^1$ function. If $f\in H^{\frac{1}{2}}(\mathbb{R})$ is a Sobelev 1/2 function that $f=f*f$, then how do you show that $f\in C^1$ with a bounded derivative. 
 A: Use the decay of $\widehat{f}$. Since $f=f\ast f$, we have that $\widehat{f}=(\widehat{f})^{2}$ and therefore
\begin{align*}
\int_{\mathbb{R}}\left(1+\left|\xi\right|^{2}\right)^{1/2}\left|\widehat{f}(\xi)\right|d\xi&=\int_{\mathbb{R}}\left(1+\left|\xi\right|^{2}\right)^{1/2}\left|\widehat{f}(\xi)\right|^{2}d\xi<\infty,
\end{align*}
which implies that $(1+\left|\xi\right|^{2})^{1/2}\widehat{f}\in L^{1}(\mathbb{R})$ and in particular, that $\widehat{f}\in L^{1}(\mathbb{R})$. This last observation shows by Fourier inversion that $f$ is equal to a continuous function a.e., which we will identify as $f$. To see that $f$ is in fact $C^{1}$, we use Fourier inversion and dominated convergence. Indeed,
\begin{align*}
\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\int_{\mathbb{R}}\widehat{f}(\xi)\dfrac{e^{2\pi i\xi(x+h)}-e^{2\pi i \xi x}}{h}d\xi
\end{align*}
Since 
$$\left|\widehat{f}(\xi)\dfrac{e^{2\pi i\xi(x+h)}-e^{2\pi i \xi x}}{h}\right|\leq C(1+\left|\xi\right|^{2})^{1/2}\left|\widehat{f}(\xi)\right|\in L^{1}(\mathbb{R}),$$
where $C>0$ is some constant, we can use dominated convergence to take the limit inside the integral and obtain that
$$f'(x)=\int_{\mathbb{R}}\widehat{f}(\xi)(2\pi i \xi)e^{2\pi i \xi x}d\xi$$
Lastly,
$$\left|f'(x)\right|\leq\int_{\mathbb{R}}\left|\widehat{f}(\xi)(2\pi i \xi)e^{2\pi i \xi x}\right|d\xi\leq\int_{\mathbb{R}}\left(1+\left|\xi\right|^{2}\right)^{1/2}\left|\widehat{f}(\xi)\right|d\xi=\left\|f\right\|_{H^{1/2}}$$
