Let $f$ a function smooth sufficiently, e.g $f\in\mathcal{S}(\mathbb{R})$ ($\mathcal{S}(\mathbb{R})$ denote the Schwartz space), and consider $0<\beta,\alpha<1$ such that $1-\beta\leqslant \alpha.$
Define the fractional laplacian by $$\Lambda^sf=(-\partial_{xx})^{\frac{s}{2}}f.$$
My question is:
$$||\Lambda^{1-\beta}f||_{L^{\infty}}\leqslant C ||f||_{C^{\alpha}}$$
where $C^{\alpha}$ is the $\alpha^{th}$-Holder norm.
Best Answer
I claim that one has the following continuous embedding property: for any $\epsilon>0$ there exists $C_{\epsilon}>0$, such that for every $f \in C^{\alpha+\epsilon}$, one has the bound $$\|(-\partial_x^2)^{\alpha/2}f\|_{L^{\infty}} \leq C\|f\|_{C^{\alpha+\epsilon}}.$$ I believe that this is optimal (i.e., $C_{\epsilon}$ blows up as $\epsilon \downarrow 0$, but that is harder, maybe try asking on mathoverflow for the sharpest possible result). I could be wrong.
Now let's prove it. Use the heat kernel representation of the fractional Laplacian: $$(-\partial_x^2)^{\alpha/2}f = C_{\alpha} \int_0^{\infty} t^{-1-\frac{\alpha}{2}}(P_tf-f)dt, \tag{1}$$ where $P_tf(x) = (p_t*f)(x)$ is the heat semigroup (i.e., $p_t(x) = \frac{1}{2\pi t}e^{-x^2/2t}$ and $"*"$ denotes convolution). Since $p_t$ integrates to $1$, one easily computes that for $\gamma>\alpha$, $$|P_tf(x)-f(x)| \leq \int_{\Bbb R}p_t(y)|f(x-y)-f(x)|dy \leq [f]_{\gamma}\int_0^{\infty} |y|^{\gamma}p_t(y)dy=Ct^{\gamma/2}[f]_{\gamma}.\tag{2}$$ On the other hand, as $P_t$ is contractive, one also has that $$\|P_tf-f\|_{L^{\infty}} \leq \|P_tf\|_{L^{\infty}} + \|f\|_{L^{\infty}} \leq 2\|f\|_{L^{\infty}}. \tag{3}$$ Next, we split the integral in (1) into two pieces, one over $t \in [0,1]$ and the other over $t \in [1,\infty)$. For the integral over $[0,1]$ we use bound (2), and for the integral over $[1,\infty)$ we use bound (3). This will give $$\|(-\partial_x^2)^{\alpha/2}\|_{L^{\infty}} \leq C[f]_{\gamma}\int_0^1 t^{\frac{\gamma}{2}-\frac{\alpha}{2}-1}dt + C'\|f\|_{L^{\infty}}\int_1^{\infty} t^{-\frac{\alpha}{2}-1}dt. $$ Since $\gamma>\alpha$, we can be sure that both integrals converge, completing the proof. It might be interesting to get the reverse embedding, now sure how though.
Edit: I guess one could also have gotten the same result by using the singular integral representation of the fractional laplacian. This would have resulted in a similar but conceptually simpler proof, splitting the integral over small $y$ and large $y$, then using $\|f\|_{L^{\infty}}$ to bound the integral over large $y$ and using $[f]_{\gamma}$ to bound the integral over small $y$.