An $f\in H^{1/2}$ with self-convolution, showing it is an $C^1$ function.

Question

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.

Answer 1

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, $$\left|f'(x)\right|\leq\int_{\mathbb{R}}\left|\widehat{f}(\xi)(2\pi i \xi)e^{2\pi i \xi x}\right|d\xi\leq\int_{\mathbb{R}}\left(1+\left|\xi\right|^{2}\right)^{1/2}\left|\widehat{f}(\xi)\right|d\xi=\left\|f\right\|_{H^{1/2}}$$