$f \in H^{s}(\mathbb{R}^{n})$, $s > n/2$, then $f$ is uniformly Holder continuous of $\alpha \in (0,1)$ order?? How do I show that if $f \in H^{s}(\mathbb{R}^{n})$, $s > n/2$, then $f$ is uniformly Holder continuous of $\alpha \in (0,1)$ order??
I know I need to show that
$$|f(x) - f(y)| \leq C\|f\|_{H^{s}}|x-y|^{\alpha} \ \ a.e. \ \ x,y\in \mathbb{R}^{n}$$
My advisor sent me to prove that:
\begin{eqnarray}\label{eq}
|f(x+h) -f(x)| \leq (2^{1-\alpha})(2\pi)^{-n/2}|h|^{\alpha}\|f\|_{H^{s}} \bigg[\int_{\mathbb{R}^{n}}d\xi(1+|\xi|^{2})^{-(s-\alpha)}\bigg]^{1/2}
\end{eqnarray}
but I have no idea how to prove the above inequality, any ideas? Excuse me :(
Remember if
$$
H^{s}(\mathbb{R}^{n}) = \lbrace f \in S^{\prime}(\mathbb{R}^{n}), (1+|\xi|^{2})^{s/2}\mathcal{F}f \in L^{2}(\mathbb{R}^{n}) \rbrace
$$
 A: Accoring to context, your Fourier transform is defined by
$$\mathcal{F}f(\xi)=\hat f(\xi)=(2\pi)^{-n/2}\int_{\mathbb R^n}f(x)e^{-ix\cdot\xi}\,dx,$$
$$ \mathcal{F}^{-1}f(x)=\check{f}(x)=(2\pi)^{-n/2}\int_{\mathbb R^n}f(\xi)e^{ix\cdot\xi}\,d\xi.$$
For a.e. $x,h\in\mathbb R^n$, we have
\begin{align*}
|f(x+h)-f(x)|&=(2\pi)^{-n/2}\left|\int_{\mathbb R^n}\hat f(\xi)e^{ix\cdot\xi}(e^{ih\cdot\xi}-1)\,d\xi\right|\\
&\leq (2\pi)^{-n/2}\int_{\mathbb R^n}|\hat f(\xi)||e^{ih\cdot\xi}-1|\,d\xi\\
&\leq (2\pi)^{-n/2}\left(\int_{\mathbb R^n}(1+|\xi|^2)^s|\hat f(\xi)|^2\,d\xi\right)^{1/2}\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-s}|e^{ih\cdot\xi}-1|^2\,d\xi\right)^{1/2}\\
&=(2\pi)^{-n/2}\|f\|_{H^s}\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-s}|e^{ih\cdot\xi}-1|^2\,d\xi\right)^{1/2}.
\end{align*}
Now we prove that for any $\gamma\in[0,1]$, we have
$$|e^{it}-1|\leq 2^{1-\gamma}|t|^\gamma,\qquad t\in\mathbb R.$$
Indeed, it is quite easy to compute that
$$|e^{it}-1|=(|\cos t-1|^2+\sin^2t)^{1/2}=(2-2\cos t)^{1/2}=2\left|\sin \frac t2\right|,\qquad t\in \mathbb R.$$
If $|t/2|\leq 1$, then we use $|\sin s|\leq |s|$ to get that $|\sin(t/2)|\leq |t/2|\leq |t/2|^\gamma$ (recall that $\gamma\in[0,1]$); if $|t/2|>1$, then $|\sin(t/2)|\leq 1<|t/2|^\gamma$. Therefore, we have checked that $|e^{it}-1|\leq 2|t/2|^\gamma=2^{1-\gamma}|t|^\gamma$ for all $t\in\mathbb R$.
We come back to estimate $|f(x+h)-f(x)|$. Since $s>n/2$, we can take $\alpha\in(0,1)$ such that $s-\alpha>n/2$. Now fix such $\alpha$. We have that $|e^{ih\cdot\xi}-1|\leq 2^{1-\alpha}|h\cdot\xi|^\alpha\leq 2^{1-\alpha}|h|^\alpha|\xi|^\alpha,$ thus
\begin{align*}
|f(x+h)-f(x)|&=(2\pi)^{-n/2}\|f\|_{H^s}\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-s}|e^{ih\cdot\xi}-1|^2\,d\xi\right)^{1/2}\\
&\leq (2\pi)^{-n/2}\|f\|_{H^s}\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-s}(2^{1-\alpha}|h|^\alpha|\xi|^\alpha)^2\,d\xi\right)^{1/2}\\
&= 2^{1-\alpha}(2\pi)^{-n/2}|h|^\alpha\|f\|_{H^s}\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-s}(|\xi|^2)^\alpha\,d\xi\right)^{1/2}\\
&\leq 2^{1-\alpha}(2\pi)^{-n/2}|h|^\alpha\|f\|_{H^s}\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-(s-\alpha)}\,d\xi\right)^{1/2}.
\end{align*}
Since $s-\alpha>n/2$, the integral $\int_{\mathbb R^n}(1+|\xi|^2)^{-(s-\alpha)}\,d\xi$ is convergent. Hence we have proved that for all $\alpha\in(0,1)$ with $s-\alpha>n/2$, there exists a constant $C=C(\alpha,s,n)$ such that
$$|f(x+h)-f(x)|\leq C\|f\|_{H^s}|h|^\alpha,\qquad \text{a.e. }\  x,h\in\mathbb R^n.$$
This implies that $f$ can be modified (a.e.) to a function which is uniformly H$\ddot{\text{o}}$lder continuous of $\alpha$ order for all $\alpha\in(0,1)\cap\left(0,s-\frac n2\right)$.
