April 18, 2011

Harmonic Analysis Lecture Notes 19

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:51 pm

Theorem (Gagliado–Nirenberg Inequality) We have

\displaystyle  	\| f\|_{L^p} \lesssim \| f\|_{L^q}^{1-\theta} \|\nabla f\|_{L^2}^\theta


\displaystyle  	\theta = \tfrac{2d(p-q)}{p\big(2d - (d-2) q\big)} 	\hbox{\hskip 18pt and \hskip 18pt} 	\left\{\begin{matrix} 	1 \leq q< p < \tfrac{d}{d-2} 		&\hbox{\hskip 18pt} d \geq 3 \\ \\ 	1 \leq q < p < \infty 			&\hbox{\hskip 18pt} d=2 \\ \\ 	1 \leq q < p \leq \infty 		&\hbox{\hskip 18pt} d = 1 	\end{matrix} \right. 	.

Proof: From Bernstein’s inequality (because {p > q})

\displaystyle  	\| f_N \|_{L^p} \lesssim \underbrace{N^{d( \frac{1}{q} - \frac{1}{p})}}_{\text{positive power of }N} \| f\|_{L^q}.

When {p \geq 2}

\displaystyle  	\| f_N \|_{L^p} \lesssim N^{d( \frac{1}{2} - \frac{1}{p})} \|f_N \|_{L^2} \lesssim \underbrace{N^{d( \frac{1}{2} - \frac{1}{p})-1}}_{\text{negative power of }N} \|\nabla f \|_{L^2}.

Moreover, by using the “wrong {p}” above (e.g. {p = 2}) and then interpolate with {L^q} (actually, genuine Hölder by letting {f \in L^q}) we see that for {q < p \leq 2},

\displaystyle  	\| f_N\|_{L^p} \lesssim \underbrace{N^{-\phi}}_{\text{negative power of }N} \|f\|_{L^q}^{1-\phi} \|\nabla f\|_{L^2}^\phi

with some {\phi \in (0,1]} determined by {\frac{1}{p} = \frac{1-\theta}{q} + \frac{\phi}{2}}.

We treat the remainder of the proof in the {p \geq 2} case—the other region is similar. Now,

\displaystyle  	\| f \|_{L^p} \leq \sum_{N \in 2^{\mathbb Z}} \|f_N\|_{L^p} \lesssim \sum_{N \in 2^{\mathbb Z}} \min \left\{ \underbrace{N^{d(\frac{1}{q} - \frac{1}{p})}}_{\text{positive power of }N} \| f\|_{L^q} , \underbrace{N^{d(\frac{1}{2} - \frac{1}{p}) - 1}}_{\text{negative power of }N} \| \nabla f\|_{L^2} \right\}.

To finish, we figure out when the minimum switches and then sum the geometric series. The details are left as an exercise. \Box

Now we ask

Question: If {\int_{{\mathbb R}^2} |\psi|^2 = N}, is

\displaystyle  	E(\psi) := \int_{{\mathbb R}^2} \frac{1}{2} |\nabla \psi|^2 - \frac{1}{4} |\psi|^4

positive, or even bounded from below? Think, stars.

Well, if {N << 1}, then we do have positivity, by Gagliardo–Nirenberg

\displaystyle  	E = \frac{1}{2} \|\nabla \psi\|_{L^2}^2 - \frac{C_{GN}^4}{4} \underbrace{\| \psi\|_{L^2}^2}_{ = N} \| \nabla \psi\|_{L^2}^2

(note that with {p = 4, q = 2}, {\theta_{GN} = \frac{1}{2}}). Thus if {N} is small

\displaystyle  	E \gtrsim \| \nabla \psi\|_{L^2}^2 \geq 0.

Conversely, if we take {\psi(x) = \lambda \psi_0(x)} with {\psi_0 \in \mathcal{S}({\mathbb R}^d)}, then

\displaystyle  	E(\psi) = \frac{1}{2} \lambda^2 \| \nabla \psi_0 \|_{L^2}^2 - \frac{1}{4} \lambda^4 \|\psi_0\|_{L^4}^4

which will be negative once {\lambda} is large enough, though this makes {N = \lambda^2 \int |\psi_0|^2} large.

Thus there is a transition size {N} at which we switch from {E > 0} always to {E <0} sometimes. What happens if we rescale {\psi} via {\psi_\lambda(x) = \lambda \psi(\lambda x)} to keep {\int |\psi|^2} constant? We obtain

\displaystyle  \begin{array}{rcl}  	\| \nabla \psi_\lambda\|_{L^2}^2 &=& \displaystyle \int_{{\mathbb R}^2} |\nabla \psi_\lambda|^2 \; dx \\ \\ 	\| \psi \|_{L^4}^4 &=& \displaystyle \int_{{\mathbb R}^2} \lambda^4 |\psi(\lambda x)|^4 \; dx = \lambda^2 \| \psi \|_{L^4}^4 	\end{array}


\displaystyle  	E(\psi_\lambda) = \lambda^2 E(\psi)

and so as soon as {\psi} has negative energy, a re-scaled version has arbitrarily negative energy.

Theorem (Rellich–Kondrachov) Let

\displaystyle  	\begin{matrix} 	1 \leq q< p < \tfrac{d}{d-2} 		&\hbox{\hskip 18pt} d \geq 3 \\ \\ 	1 \leq q < p < \infty 			&\hbox{\hskip 18pt} d=2 \\ \\ 	1 \leq q < p \leq \infty 		&\hbox{\hskip 18pt} d = 1 	\end{matrix}.

Then for all {R > 0},

\displaystyle  	\{ \chi_R f : \| f \|_{L^q} \leq 1 \hbox{ and } \| \nabla f \|_{L^2} \leq 1 \}

is compact in {L^p}, where {\chi_R = \chi_{B(0,R)}}.

Remark Note that, without the {\chi}, this result would be false. For example, consider {\{ \phi(x - x_0), \forall x_0 \in {\mathbb R}^d\}} with fixed {\phi \in \mathcal{S}({\mathbb R}^d)}.

Proof: From the previous proof, we know that

\displaystyle  	\| f_N \|_{L^p} \lesssim N^{-\delta},\hbox{\hskip 18pt} \delta(p,q,d) > 0.

and thus

\displaystyle  	\| f_{\geq N} \|_{L^p} \lesssim \sum_{M \geq N} \|f_M\|_{L^p} \lesssim N^{-\delta}.

Therefore, the question reduces to showing that

\displaystyle  	\{ \chi_R f_{\leq N} : \| f \|_{L^q} \leq 1 \hbox{ and } \| \nabla f \|_{L^2} \leq 1 \}

is compact for all {N}. There are many approaches to prove this.

The first approach is to recall that a {\mathcal{F} \subseteq L^p} is pre-compact if and only if {\mathcal{F}} is

  1. Bounded: {\exists C > 0} such that for all {f \in \mathcal{F}}, we have {\|f\|_{L^p} \leq C}.
  2. Tight: {\forall \epsilon > 0 \exists R >0} such that for all {f \in \mathcal{F}}, we have {\|f\|_{L^p(|x| > R)} < \epsilon}.
  3. Equicontinuous: {\forall \epsilon > 0 \exists \delta > 0} such that for all {|y| < \delta} and for all {f \in \mathcal{F}}, we have {\int |f(x+y) - f(x)|^o \; dx < \epsilon}.

In our case, property 1 follows by the Gagliardo–Nirenberg inequality, property 2 follows by the analysis of {\chi_R}, and property 3 follows by the analysis of {P_{\leq N} f = N^d \check \phi(N \cdot) * f}.

The second approach is to note that, because we are on a compact/bounded region (namely, {\overline{B(0,R)}}), compactness of {\{\chi_R f_{\leq N}\}} as a set of continuous functions in the supremum norm implies compactness in {L^p}. By Arzelà–Ascoli, we need only to check boundedness and equicontinuity:

\displaystyle  	\|f_{\leq N}\|_{L^\infty} \lesssim N^{d/q}, \hbox{ and } \|f\|_{L^q} , \|\nabla f_{\leq N} \|_{L^\infty} \leq N^{1 + \frac{d}{q}} \|f\|_{L^q}.

which follow by Bernstein’s inequalities. \Box

Theorem (Radial Gagliardo–Nirenberg) If {f \in L^q \cap \dot H^1} and is spherically symmetric, then

\displaystyle  	|x|^\frac{2(d-1)}{q+2} |f(x)| \lesssim \|f\|^{\frac{q}{q+2}}_{L^q} \| \nabla f \|_{L^2}^\frac{2}{q+2}

Proof: Let us write {f} as a function of radius {r}. By the fundamental theorem of calculus,

\displaystyle  	(f(r))^{1+\frac{q}{2}} = -\int_r^\infty (1+\frac{q}{2}) |f(\rho)|^{\frac{q}{2} - 1} \bar f(\rho) f'(\rho) \; d\rho.


\displaystyle  \begin{array}{rcl}  	|f(r)|^{1+\frac{q}{2}} &\underbrace{\lesssim}_{\rho > r}& \displaystyle r^{-(d-1)} \int_r^\infty \rho^\frac{d-1}{2} |f(\rho)|^\frac{q}{2} \rho^\frac{d-1}{2} |f'(\rho)| \; d\rho \\ \\ 	\scriptsize\hbox{[Cauchy--Schwarz] \hskip 18pt}	&\lesssim& r^{-(d-1)} \sqrt{ \int_0^\infty |f|^q \rho^{d-1} \; d\rho} \sqrt{ \int_0^\infty |f'(\rho)|^2 \; \rho^{d-1} \; d\rho}, 	\end{array}


\displaystyle  	|f(x)|^\frac{2+q}{2} \lesssim |x|^{-(d-1)} \|f\|_{L^q}^\frac{q}{2} \|\nabla f \|_{L^2}.


Corollary The set

\displaystyle  	\{ f : \|f\|_{L^q} \leq 1, \|\nabla f\|_{L^2} \leq 1, \hbox{ and } f \hbox{ is spherically symmetric}\}

is compact in {L^p} when {d \geq 2}, where {p} and {d} are as above.

Remark This result is false when {d = 1} (c.f. two bump functions who are symmetric with respect to each other about 0 and letting them move to infinity).

Proof: We just need to check tightness, so that we can apply Rellich–Kondrachov, which we can do via interpolation:

\displaystyle  	\|f\|_{L^p(|x| > R)} \lesssim \|f\|_{L^q(|x| > R)}^\theta \underbrace{\|f\|_{L^\infty(|x| \geq R)}^{1-\theta}}_{\|f\|_{L^\infty(|x| > R)} \lesssim R^{-\delta} \|f\|_{L^q}^\phi \|\nabla f\|_{L^2}^{1-\phi}}.


Finally, this leads us to

Theorem (Sharp Gagliardo–Nirenberg) Let {d \geq 2} and {2 < p < \frac{2d}{d-2}}. The function

\displaystyle  	J(f) = \frac{\|f\|_{L^p}}{\|f\|_{L^2}^{1-\theta} \|\nabla f\|_{L^2}^{\theta}}, \hbox{\hskip 18pt where \hskip 18pt} \theta = \frac{(p-2)d}{2p}

achieves its maximum on {H^1({\mathbb R}^d)\setminus\{0\} = \{ f \not \equiv 0 : f \in L^2, \nabla f \in L^2\}}. Moreover, any such maximizer can be transformed via {g(x) = \alpha f(\beta x)} to a solution of

\displaystyle  	-\Delta g - |g|^{p-2} g = -g

In particular, this partial differential equation has a solution.

Remark One can show (we will not do so here) that there is a unique radial non-negative solution {Q} to this PDE in {H^1}. Moreover, one can deduce that every optimizer has the form {f = \alpha Q(\beta(x - x_0))} and even that the Hessian at these points is positive definite perpendicular to the minimizing manifold.

Proof: By the Gagliardo–Nirenberg inequality, {J} is bounded. Let {f_n} be a sequence in {H^1} so that {J(f_n)} converges to the extremal value. Replacing {f_n} by their radial symmetric decreasing rearrangements {f_n^*} will only increase (or retain) the value of {J}, hence {f_n^*} is also an optimizing sequence.

Next we consider replacing {f_n^*(x)} by {\alpha_n f_n^*(\beta_n x)}. It is easy to check that this does not affect the value of {J}. We choose {\alpha_n, \beta_n} so that {\|\alpha_n f_n^*(\beta_n x)\|_{L^2} = \| \nabla (\alpha_n f_n^*(\beta_n x))\|_{L^2} = 1}. Henceforth, we call {\alpha_n f_n^*(\beta_n x) = f_n}.

By Alaoglu, we can pass to a subsequence so that {f_n \stackrel{\text{weak-}*}{\rightharpoonup} f} in {L^2} and {\nabla f_n \rightharpoonup F}, also weak-{*} in {L^2}. Note that since {\nabla f} is defined via testing against {\phi \in C_c^\infty}, we are guaranteed that {F = \nabla f}. Lower semi-continuity of norms under weak-{*} limits shows that {\|f\|_{L^2} \leq \liminf \|f_n\|_{L^2} = 1} and, similarly, that {\| \nabla f \|_{L^2} \leq 1}. [Note that Alaoglu in {L^p} leads nowhere.]

By Rellich–Kondrachov and tightness for radial functions, we know that a further subsequence converges strongly in {L^p}. [Actually, we do not need to pass to a further subsequence because testing with {\phi \in C_c^\infty} identifies all subsequential limits as being {f}—the weak-{*} {L^2} limit.] Note that as {\|\nabla f_n\|_{L^2} = 1 = \|f_n\|_{L^2}}, we thus have {J(f_n) = \|f_n\|_{L^p} \rightarrow \|f\|_{L^p}}, which is not zero as {(\sup J) \neq 0}. Noting that

\displaystyle  	\sup J \geq \frac{\|f\|_{L^p}}{\|f\|_{L^2}^{1-\theta} \|\nabla f\|_{L^2}^\theta} \geq \|f\|_{L^p} = \sup J,

therefore {\|f_n\|_{L^2} \rightarrow \|f\|_{L^2}} and {\|\nabla f_n\|_{L^2} \rightarrow \|\nabla f\|_{L^2}} ({f} cannot beat the optimal value). By the Radon–Riesz theorem, we see that {f_n \rightarrow f} in {H^1}.

Now, let {f} be an optimizer such that {\|f\|_{L^2} = \| \nabla f \|_{L^2} = 1}. Consider {J(f+t\phi)} for some {\phi \in C_c^\infty}. This must have {\frac{d}{dt} \big|_{t=0} J(f+t\phi) = 0}. Furthermore

\displaystyle  	\partial_t \big|_{t=0} \int |f + t\phi|^p \; dx = p \text{ Re} \int |f|^{p-2} f \bar \phi \; dx


\displaystyle  \begin{array}{rcl}  	\displaystyle \partial_t |_{t=0} \int |\nabla (f + t\phi)|^2 \; dx &=& \partial_t \big|_{t=0} \int |\nabla f|^2 + 2t \text{ Re } \nabla f \cdot \nabla \bar \phi + t^2 |\nabla \phi|^2 \; dx \\ \\ 		&=& \displaystyle \int -2 \text{ Re } (\underbrace{\Delta f}_{\scriptsize\begin{matrix}\text{distributional} \\ \text{Laplacian}\end{matrix}} \phi) \; dx. 	\end{array}

Therefore, using the fact that {\|f\|_{L^2} = \| \nabla f \|_{L^2} = 1},

\displaystyle  \begin{array}{rcl}  	0 &=& \displaystyle \frac{d}{dt} \big|_{t=0} J(f+t\phi) \\ 		&=& \displaystyle \| f \|_{L^p}^{1-p} \text{ Re } \int |f|^{p-2} f \bar \phi \; dx-\|f\|_{L^p} (1 - \theta) \text{ Re } \int f \bar \phi \; dx + \theta \|f\|_{L^p} \text{ Re } \int \Delta f \bar \phi \; dx, 	\end{array}

valid for every {\phi \in C_c^\infty}. Choosing {\phi} and {i\phi} and canceling {\| f\|_{L^p}}, we obtain

\displaystyle  	\int \big( \theta \Delta f - (1 - \theta) f + \|f\|_{L^p}^{-p} |f|^{p-2} f \big) \bar\phi \; dx = 0

for all {\phi \in C_c^\infty}. Thus {f} is a distributional solution to the PDE

\displaystyle  	\theta \nabla f + (\max J)^p |f|^{p-2} f = (1-\theta) f

which is called the Euler–Lagrange equation for our extremal problem. By rescaling {g(x) = \tilde \alpha f(\tilde \beta x)}, with {\tilde \alpha, \tilde \beta} determined by {\theta, \max J} alone, we get

\displaystyle  	\Delta g + |g|^{p-2} g = g

To recap, every optimizer will, after suitable rescaling, solve the above PDE. Moreover, there is at least one optimizer that has {f = f^*}. \Box

As a sketch of “uniqueness” of optimizers: given an optimizer {f}, note that {f^*} is also an optimizer. Moreover, we must have {\int |\nabla f|^2 = \int |\nabla f^*|^2}. In general, this is not enough to guarantee that {f(x) = f^*(x-x_0)} (see submarine remark above); however this conclusion does follow if we can show that the set {\{f^*=\lambda\}} has zero measure for {\lambda >0}. Switching to polar coordinates, we see that radial optimizers obey an ODE (in a distributional sense)

\displaystyle  	\partial_r^2 f^* + \frac{d-1}{r} \partial_r f^* + |f^*|^{p-2} f^* = f^*

after rescaling. As such, we can see that {f^*} is actually real analytic (at least away from {r \equiv 0}) and so has no positive measure level set — we can also realize this from the fact that {f^*} radially decreasing means that a plateau would force our ODE to have a locally constant solution, but uniqueness for ODES implies this would be a globally constant solution {1}, which fails to be in {H^1}. Lastly, we need the uniqueness of non-negative {H^1} solutions to our ODE.

The proof failed to cover the case {d = 1} and {2 < p < \infty} because {\{ f \text{ even }: \|f\|_{H^1} \leq 1\}} is not tight in {L^p} — for example, consider {f = \phi(x - n) + \phi(x + n)}. Nevertheless, {\{ f^* : \|f\|_{H^1} \leq 1\}} is indeed tight in {L^p}, which allows us to adapt the preceding proof.

Proof of tightness: We compute

\displaystyle  	1 \geq \int |f^*(x)|^2 \; dx \geq \int_{-R}^R |f^*(x)|^2 \; dx \geq 2R \cdot |f^*(R)|^2

because {f^*} is decreasing in {|x|}. Thus

\displaystyle  	\| f^* \|_{L^\infty(|x| \geq R)} = |f^*(R)| \leq \frac{\| f\|_{L^2}}{\sqrt{2R}}

and so by Hölder’s inequality

\displaystyle  	\|f^*\|_{L^p(|x| \geq R)} \leq \|f\|_{L^2}^\frac{2}{p} \|f\|_{L^\infty(|x| \geq R)}^{1-\frac{2}{p}} \leq \frac{\|f\|_{L^2}}{(2R)^\frac{p-2}{2p}}. \qed


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

Blog at WordPress.com.

%d bloggers like this: