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## April 18, 2011

### Harmonic Analysis Lecture Notes 19

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

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

provided

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

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. $\Box$

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

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

Therefore

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

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,

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

Therefore,

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

namely,

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

$\Box$

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:

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

$\Box$

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

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

and

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

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

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