Relation between Holder norm and fractional laplacian 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.
 A: 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 $y$.
A: For completeness, we will prove a partial converse to the bound given in the other answer. Specifically, we show that for all $\epsilon>0$ and compact sets $K \subset \Bbb R$ there exists $C=C_{\epsilon,K}$ such that for all sufficiently smooth $f$ one has $$\sup_{x,y \in K}\frac{|f(x)-f(y)|}{|x-y|^{\alpha-\epsilon}} \leq C\|(-\partial_x^2)^{\alpha/2}f\|_{L^{\infty}}.$$
To show this let $g = (-\partial_x^2)^{\alpha/2}f$. We use the heat kernel representation of the inverse operator to the fractional laplacian: $$(-\partial_x^2)^{-\alpha/2}g= C_{\alpha} \int_0^{\infty} t^{\frac{\alpha}2-1}P_tg\;dt,\tag{1}$$where our notations are the same as in the other answer.
Note that $$|P_tg(x)-P_tg(y)| \leq \int_0^{\infty} |g(z) ||p_t(x-z)-p_t(y-z)|dz \leq \|g\|_{L^{\infty}} \int_0^{\infty}|p_t(x-z)-p_t(y-z)|dz.$$ Now, using $|p_t(x-z)-p_t(y-z)| \leq p_t(x-z)+p_t(y-z)$ one obtains $$|P_tg(x)-P_tg(y)| \leq C\|g\|_{L^{\infty}}. \tag{2}$$ On the other hand, if we use the bound $|p_t(x-z)-p_t(y-z)| \leq Ct^{-3/2} |x-y| \big(|x-z|+|y-z|\big)e^{-|x-z|^2\wedge |y-z|^2/2t}$, then one obtains $$|P_tg(x)-P_tg(y)| \leq C\|g\|_{L^{\infty}}|x-y|t^{-1/2}. \tag{3}$$ By geometrically interpolating between (2) and (3) we see that for any $\gamma \in[0,1]$ one has that $$|P_tg(x)-P_tg(y)| \leq C\|g\|_{L^{\infty}}|x-y|^{\gamma}t^{-\gamma/2}. \tag{4}$$ Now we are going to write $|(-\partial_x^2)^{-\alpha/2}g(x)-(-\partial_x^2)^{-\alpha/2}g(y)|$ as an integral according to (1). Then we split the integral into two pieces, one over $t\in[0,1]$ and the other over $t\in[1,\infty)$. For the integral over $[0,1]$ we will use the bound (4) with any $\gamma<\alpha$, and over $[1,\infty)$ we use (4) with $\gamma=1$. This will give $$|(-\partial_x^2)^{-\alpha/2}g(x)-(-\partial_x^2)^{-\alpha/2}g(y)| \leq C\|g\|_{L^{\infty}}|x-y|^{\gamma}\int_0^1 t^{\frac{\alpha}{2}-\frac{\gamma}{2}-1}dt+C\|g\|_{\infty}|x-y|\int_1^{\infty} t^{-\frac{\alpha}{2}-\frac{3}{2}}dt,$$ proving the claim since $\gamma<\alpha\le 1$ ensures that all integrals converge. Again, we could have done a singular integral characterization of $(-\partial_x^2)^{-\alpha/2}$ but we chose heat kernel instead.
