# Some embedding results

Definition ${X,Y}$ are Banach spaces, we say ${X}$ is embedded in ${Y}$, denoted by ${X \hookrightarrow Y}$, if ${X \subset Y}$ and ${\| \cdot \|_Y \le C \| \cdot \|_X}$.

Notation: ${\langle \xi \rangle := \sqrt{1+\xi^2}}$.

Here are some embedding results.

Theorem 1 ${H^s({\mathbb R}^n) \hookrightarrow C_0({\mathbb R}^n)}$, if ${s > \frac{n}{2}}$.

Proof:

$\displaystyle \|f\|_{L^\infty} \le C \| \hat{f} \|_{L^1} = C \| \langle \xi \rangle ^s \hat{f} \langle \xi \rangle ^{-s}\|_{L^1} \le C \| \langle \xi \rangle ^s \hat{f} \|_{L^2} \| \langle \xi \rangle ^{-s}\|_{L^2}$

$\displaystyle = C \|f \|_{H^s} \left( \int_{{\mathbb R}^n} \frac{1}{(1+\xi^2)^s} \, d \xi \right)^{\frac{1}{2}} \le C \|f \|_{H^s}$

since ${2s > n}$.

If ${f \in H^s({\mathbb R}^n)}$, then ${\hat{f} \in L^1({\mathbb R}^n)}$. Therefore ${f}$ is continuous and vanishes at ${\infty}$ (Riemann Lebesgue Theorem). $\Box$

Theorem 2 ${H^s({\mathbb R}^n) \hookrightarrow H^t({\mathbb R}^n)}$, if ${s \ge t \ge 0}$.

Proof:

$\displaystyle \|f \|_{H^t} = \| \langle \xi \rangle ^t \hat{f} \|_{L^2} \le \| \langle \xi \rangle ^s \hat{f} \|_{L^2} = \|f \|_{H^s}.$

$\Box$

Theorem 3 (Sobolev embedding) If ${0 \le s < \frac{n}{2}}$, then ${H^s({\mathbb R}^n) \hookrightarrow L^p({\mathbb R}^n)}$ for ${p = \frac{2n}{n-2s}}$.

Proof: The idea is to split ${f}$ into low frequency part and high frequency part in Fourier space, i.e.

$\displaystyle f = \mathcal{F}^{-1} \left( \chi_{B(0,N)} \hat{f} \right) + \mathcal{F}^{-1} \left( \chi_{B(0,N)^\mathsf{c}} \hat{f} \right) =: f_N^L + f_N^H,$

where ${\mathcal{F}^{-1}}$ is the inverse Fourier transform. Note that

$\displaystyle \|f_N^L\|_{L^\infty} \le C\| \hat{f}_N^L \|_{L^1} = C \int_{|\xi| < N} |\hat{f}(\xi)| \,d \xi \le C \|f \|_{H^s} \int_{|\xi| < N} \langle \xi \rangle ^{-2s} \,d \xi \le C \|f \|_{H^s} N^{\frac{n}{2} -s}.$

So if we define

$\displaystyle N_\lambda := \left( \frac{\lambda}{2C \|f \|_{H^s}} \right)^ {\frac{2}{n-2s}} \quad \Leftrightarrow \quad C \|f \|_{H^s} N_\lambda^{\frac{n}{2} -s} = \frac{\lambda}{2} \quad \Rightarrow \quad \|f_{N_\lambda}^L\|_{L^\infty} \le \frac{\lambda}{2}.$

Since

$\displaystyle \left\{ |f| >\lambda \right\} \subset \left\{ |f_{N_\lambda}^L| > \frac{\lambda}{2} \right\} \cup \left\{ |f_{N_\lambda}^H| > \frac{\lambda}{2} \right\} = \left\{ |f_{N_\lambda}^H| > \frac{\lambda}{2} \right\},$

we have

$\displaystyle \left| \left\{ |f| >\lambda \right\} \right| \le \left| \left\{ |f_{N_\lambda}^H| > \frac{\lambda}{2} \right\} \right| \le \frac{4}{\lambda^2} \|f_{N_\lambda}^H\|_{L^2}^2 = \frac{4}{\lambda^2} \int_{|\xi|>N_\lambda} |\hat{f}(\xi)|^2 \, d \xi.$

Then we know

$\displaystyle \|f\|_{L^p}^p = p \int_0^\infty \lambda^{p-1} \left| \left\{ |f| >\lambda \right\} \right| \,d \lambda \le 4p \int_0^\infty \lambda^{p-3} \int_{|\xi|>N_\lambda} |\hat{f}(\xi)|^2 \, d \xi d \lambda.$

Since the integrand is nonnegative, we can use Tonelli theorem to change the order of integration and get

$\displaystyle \|f\|_{L^p}^p \le 4p \int_{{\mathbb R}^n} |\hat{f}(\xi)|^2 \int_0^{C \|f \|_{H^s} |\xi|^{\frac{n}{2} -s}} \lambda^{p-3} \, d\lambda d \xi = \frac{4pC}{p-2} \int_{{\mathbb R}^n} |\hat{f}(\xi)|^2 \|f \|_{H^s}^{p-2} |\xi|^{\frac{n-2s}{2} \cdot (p-2)}\, d\xi$

$\displaystyle =C \|f \|_{H^s}^{p-2} \int_{{\mathbb R}^n} |\hat{f}(\xi)|^2 |\xi|^{2s} \, d\xi = C\|f \|_{H^s}^p.$

$\Box$

Theorem 4 (Moray Inequality) If ${s > \frac{n}{2}}$, then ${H^s({\mathbb R}^n) \hookrightarrow C^\alpha({\mathbb R}^n)}$, for ${\alpha = s - \frac{n}{2}}$.

Proof: Take an ${f \in H^s({\mathbb R}^n)}$ and first assume ${\alpha = s - \frac{n}{2} \in (0,1]}$. By Theorem 1 we know ${f \in L^\infty}$, so we need to check ${|f(x) - f(y)| \le C\| f \|_{H^s} |x-y|^\alpha}$ for any ${x,y}$.

$\displaystyle |f(x)-f(y)| = \left| \frac{1}{(2\pi)^{n/2}} \int \left( e^{ix \xi} - e^{iy \xi} \right) \hat{f}(\xi)\, d\xi \right| \le C \int \min (|x-y|\cdot |\xi| ,1) |\hat{f}(\xi)| \, d\xi$

$\displaystyle \le C \left( \int \langle \xi \rangle^{2s} |\hat{f}(\xi)|^2 \, d\xi \right)^{\frac{1}{2}} \left( \int \langle \xi \rangle^{-2s} \min (|x-y|^2 |\xi^2| ,1) \, d\xi \right)^{\frac{1}{2}}$

we can split the later term into two parts, which is

$\displaystyle \int \langle \xi \rangle^{-2s}\min (|x-y|^2 |\xi^2| ,1) \, d\xi = \int_{| \xi| < \frac{1}{|x-y|}} |x-y|^2 |\xi|^2 \langle \xi \rangle^{-2s} \, d \xi + \int_{| \xi| > \frac{1}{|x-y|}} \langle \xi \rangle^{-2s} \, d \xi$

$\displaystyle \le C |x-y|^2 (|x-y|^{-1})^{n+2-2s} + C |x-y|^{2s-n} \le C |x-y|^{2s-n}$

Therefore

$\displaystyle |f(x)-f(y)| \le C \| f \|_{H^s} |x-y|^{2s-n}$

for ${\alpha =s - \frac{n}{2} \in (0,1]}$. To conclude the case where ${\alpha >1}$, we simply just use the fact that

$\displaystyle \| f \|_{C^\alpha} \sim \| \nabla f \|_{C^{\alpha-1}} + \| f \|_{L^\infty}$

$\displaystyle \| f \|_{H^s} \sim \| \nabla f \|_{H^{s-1}} + \| f \|_{L^2}.$

$\Box$