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

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