## Category Archives: Harmonic Analysis

### One example of blowing up corner

Consider ${u(x,y)=\sqrt{x^2+y^4}}$ on ${\mathbb{R}^2_+=\{(x,y):x\geq 0, y\geq 0\}}$. Notify ${\mathbb{R}^2_+}$ has a corner at the origin and ${u}$ is not smooth at the origin. ${u\approx x}$ when ${x\geq y^2}$ and ${u\approx y^2}$ when ${x\leq y^2}$. We want to resolve ${u}$ by blowing up the origin through a map ${\beta}$. After blowing up, ${W}$ looks like the following picture.

Denote ${W=[\mathbb{R}^2_+,(0,0)]}$ and ${\beta:W\rightarrow \mathbb{R}^2_+}$ is the blow down map. On ${W\backslash lb}$(near A), ${\beta}$ takes the form

$\displaystyle \beta_1(\xi_1,\eta_1)=(\xi_1^2,{\xi_1\eta_1})$

where ${\xi_1}$ is the boundary defining function for ff and ${\eta_1}$ is boundary defining function for rb. Similarly on ${W\backslash rb}$(near B), ${\beta}$ takes the form

$\displaystyle \beta_2(\xi_2,\eta_2)=(\xi_2\eta^2_2,\eta_2)$

where ${\xi_2}$ is a bdf for lb and ${\eta_2}$ is a bdf for ff. One can verify that ${\beta}$ is a diffeomorphism ${\mathring{W}\rightarrow \mathring{\mathbb{R}}^2_+}$. Let ${w=\beta^* u}$. Then ${w}$ is a polyhomogeneous conormal function on ${W}$. Its index can be denoted ${(E,F,H)}$ correspond to lb, ff and rb.

$\displaystyle E=\{(n,0)\}, F=\{(2n,0)\}, H=\{(2n,0)\}$

Suppose ${\pi_1}$ is the projection to ${x}$ coordinate. Consider ${f=\pi_1\circ \beta:W\rightarrow \mathbb{R}_+}$, ${f}$ is actually a ${b-}$fibration. Then the push forward map ${f_*}$ maps ${w}$ to a polyhomogeneous function on ${\mathbb{R}^+}$.

$\displaystyle f_*w=\pi_* u=\int_0^\infty u(x,y)dy$

In order to make ${u}$ is integrable, let us assume ${u}$ support ${x\leq 1}$ and ${y\leq 1}$. What is the index for ${f_* w}$ on ${\mathbb{R}^+}$?

$\displaystyle \int_0^1\sqrt{x^2+y^4}dy=\int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy+\int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy$

For the first integral, letting ${y^2/x=t}$

$\displaystyle \int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy=x\sqrt{x}\int_0^1\sqrt{1+t^4}dt=c_0x\sqrt{x}$

For the second integral, letting ${x/y^2=t}$

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=\frac{1}{2}x\sqrt{x}\int_x^1t^{-\frac{5}{2}}\sqrt{t^2+1}dt$

Since the Taylor series

$\displaystyle \sqrt{1+t^2}=1+\frac{1}{2}t^2-\frac{1}{8}t^4+\cdots,\quad \text{for }|t|<1$

Consequently

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=a_0+a_1x+a_2x\sqrt{x}+\cdots$

Combining all the above analysis, the index for ${f_*w}$ is ${\{(n,0)\}\cup \{\frac{n}{2},0\}}$

From another point of view, the vanishing order of ${f}$ on each boundary hypersurface of ${W}$ are ${e_f(lb)=1}$, ${e_f(\text{ff})=2}$ and ${e_f(rb)=0}$. ${f}$ maps lb and ff to the boundary of ${\mathbb{R}^+}$. Therefore the index of ${f_*w}$ is contained in

$\displaystyle \frac{1}{e_f(lb)}E\overline{\cup}\frac{1}{e_f(\text{ff})}F=E\overline{\cup}\frac{1}{2}F$

Remark: Daniel Grieser, Basics of ${b-}$Calculus.

### Compensated compactness

Suppose $T$ is a vector field and $\nabla\cdot T = 0$. $E= \nabla \psi$ and $\psi$ is a scalar function. We have following theorem(Coifman-Lions-Meyers-Semmes)

Theorem: If $T\in L^2(\mathbb{R}^n)$ and $T\in L^2(\mathbb{R}^n)$, then $E\cdot T\in \mathcal{H}^1(\mathbb{R}^n)$, which is the hardy space.
Given $f(x)\in L^1(\mathbb{R}^n)$, it has harmonic extension $\mathbb{R}^{n+1}_+=\{(x,t)|x\in\mathbb{R}^n, t>0\}$

$\tilde{f}(x,t)=c_n\int_{\mathbb{R}^n}\frac{ f(x-y)t}{(t^2+|x|^2)^{\frac{n+1}{2}}}dy$

Definition: the non-tangential maximal function

$N(f)=\sup_{(\xi,t)\in \Gamma(x)}|\tilde f(\xi, t)|$

It is easy to prove that $N(f)\leq c_n f^*(x)$ the Hardy-Littlewood maximal function. From this we can Hardy norm as

$||f||_{\mathcal{H}^1}=||f||_{L^1}+||N(f)||_{L^1}$

Hardy space consists of all $f$ having finite hardy norm. There is well know fact that the dual space of $\mathcal{H}^1$ is BMO, which is defined as the following.

Define $f\in L^1_{loc}(\mathbb{R}^n)$, if for any cube $Q$,

$\sup_Q\frac{1}{|Q|}\int_Q|f-f_Q|<\infty,\quad \text{where }f_Q=\frac{1}{|Q|}\int_Qf$

then $f\in BMO$. $L^\infty \subset BMO$ and $\log|x|\in BMO$ but not in $L^\infty$.

Let us see how do we use the main theorem. Suppose on $\mathbb{R}^2$, $u$ is the solution of the following elliptic equation

$\displaystyle\frac{\partial}{\partial x_i}\left(a_{ij}(x)\frac{\partial u}{\partial x_j}\right)=\frac{\partial f}{\partial x_1}\frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2}\frac{\partial f}{\partial x_2}$

where $||\nabla f||_{L^2}<\infty$, $||\nabla g||_{L^2}<\infty$ and $(a_{ij})$ is uniform elliptic. YanYan Li and Sagun Chanillo proved that the green function of this elliptic operator belongs to BMO. The right hand side of this equation can be rewritten as $T\cdot E$, where

$T=\left(\frac{\partial f}{\partial x_2}, -\frac{\partial f}{\partial x_1}\right),\quad E=\left(\frac{\partial g}{\partial x_1},\frac{\partial g}{\partial x_a}\right)$

therefore the right hand side belong to $\mathcal{H}^1$. Since

$u(x)=\int G_x(y)T\cdot E(y)dy$

therefore from the theorem we stated at the beginning, we get

$||u||_\infty\leq C||\nabla f||_{L^2}||\nabla g||_{L^2}$

### A rigorous way to define the singular integral

Suppose $K(x)$ satisfies the following conditions

(1)$|K(x)|\leq \frac{C}{|x|^n}$

(2)$|\nabla K(x)|\leq frac{C}{|x|^{n+1}}$

(3)$\displaystyle \int_{r_1<|x|

Then we can define the singular integral operator

$\displaystyle Tf=\int_{\mathbb{R}^n}K(x-y)f(y)dy$

On the face of it, this integrand may not be Lebesgue integrable. There is a way to interpret this integral

$\textbf{Thm: }$ Suppose (1),(2),(3) are satisfied, define

$\displaystyle T_{\epsilon,R}f(x)=\int_{\epsilon<|x-y|

Then $||Tf||_2\leq C||f||_2$ with $C$ independent of $\epsilon,R,f$. Moreover, one can prove $T_{\epsilon,R}f$ is a cauchy sequence, then there exist $g\in L^2(\mathbb{R}^n)$, such that

$\displaystyle T_{\epsilon,R}f\to g$ in $L^2$

From this, we define $Tf=g$.

### Partial sum operator of fourier series and Hilbert transform

$\mathbf{Problem:}$ Let $D_N=\sum\limits_{-N}^{N}e^{inx}$, the partial sum operator of fourier series can be represented by

$\displaystyle S_Nf(x)=\int_{-\pi}^\pi f(t)D_N(x-t)dt=f\ast D_N$

actually $\displaystyle D_N=\frac{\sin(N+\frac{1}{2})x}{\sin\frac{x}{2}}$

$\displaystyle S_Nf(x)=\int_{-\pi}^\pi f(t)\frac{\sin(N+\frac{1}{2})(x-t)}{\sin\frac{x-t}{2}}$

$\displaystyle =\int_{-\pi}^\pi f(t)\frac{\cos(N+\frac{1}{2})x\sin(N+\frac{1}{2})t-\sin(N+\frac{1}{2})x\cos(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}$

$\displaystyle =P.V.\cos(N+\frac{1}{2})x\int_{-\pi}^\pi\frac{f(t)\sin(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}+P.V.\sin(N+\frac{1}{2})x\int_{-\pi}^\pi\frac{f(t)\cos(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}$

$\displaystyle =P.V.\frac{1}{\sin\frac{t}{2}}\ast f_N^1+P.V.\frac{1}{\sin\frac{t}{2}}\ast f_N^2$

Since $\displaystyle |\frac{1}{\sin\frac{t}{2}}-\frac{1}{\frac{t}{2}}|\leq ct$, for $t\in (-\pi,\pi)$, then $S_Nf$ behavior much like Hilbert transform

$\displaystyle Hf(x)=P.V.\int_{-\pi}^\pi\frac{f(x)}{x-t}$

$\mathbf{Thm:}$

(a) $||Hf||_p\leq c_p||f||_p$, $1

(b) $\{x\in(-\pi,\pi)||Hf(x)|>\lambda\}\leq \frac{C||f||_1}{\lambda}$

By the fact in this post, $S_Nf\to f$ in $L^p(-\pi,\pi)$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### Partial sum operator of fourier series and L^p boundedness

Suppose $f\in L^1(-\pi,\pi)$ , consider the fourier series of $f$

$\displaystyle f\sim \sum \limits_{n\in\mathbb{Z}}c_ne^{inx}$

and its partial sum $\displaystyle S_N(f)=\sum \limits_{-N}^Nc_ne^{inx}$

$\mathbf{Problem:}$ The following statements are equivalent

(a)   $||S_Nf||_p\leq c_p||f||_p$, $1

(b)  $||S_Nf-f||_p\to \infty$ as $N\to \infty$

(c)   $||S_Nf||_{p'}\leq ||f||_{p'}$, $\frac{1}{p}+\frac{1}{p'}=1$

$\mathbf{Proof:}$

(a)$\Rightarrow$(b). Since trigonometric polynomial are dense in $L^p$, then we can choose $g$ to be a trigonometric polynomial are sufficiently close to $f$ in $L^p$.

$\displaystyle ||S_Nf-f||_p\leq ||S_Nf-S_Ng||_p+||S_Ng-g||_p+||g-f||_p$

$\displaystyle = ||S_N(f-g)||_p+||S_Ng-g||_p+||g-f||_p$

$\displaystyle \leq c_p||f-g||_p+||S_Ng-g||_p+||g-f||_p\to \infty$

(b)$\Rightarrow$(a). Uniform boundedness principle.

(a)$\Rightarrow$(c) and (c)$\Rightarrow$(a) are quite easy.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### Poincare inequality from different approach

$\mathbf{Problem:}$ Let $\Omega$ be convex and $u\in C^1(\Omega)$. Suppose $S\subset\Omega$ with $\displaystyle \frac{|S|}{|\Omega|}>0$,

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy$

where $d$ is the diameter of $\Omega$ and $\displaystyle u_S=\frac{1}{|S|}\int_S u(y)dy$.

$\mathbf{Proof\,1: }$

$\displaystyle u(y)-u(x)=\int_0^1\frac{d}{dt}\left(u(ty+(1-t)x)\right)dt=\int_0^1\nabla u(ty+(1-t)x)\cdot(x-y)dt$

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S|u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^1|\nabla u(ty+(1-t)x)||x-y|dtdy$

Set $z=ty+(1-t)x$, then $t|x-y|=|z-x|$. Since $|x-y|\leq d$, $\displaystyle t\geq \frac{|z-x|}{d}$. We also have $t^ndy=dz$

Then

$\displaystyle |u(x)-u_S|\leq \frac{d}{|S|}\int_\Omega\int_\frac{|z-x|}{d}^\infty \frac{|\nabla u(z)|}{t^n}\,dtdz=\frac{d^n}{(n-1)|S|}\int_\Omega\frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$.

$\mathbf{Proof\,2: }$ Using

$\displaystyle u(x)-u(y)=\int_{0}^{|x-y|}\frac{d}{dt}u(x+tw)dt=\int_0^{|x-y|}\nabla u(x+tw)\cdot wdt$

where $\displaystyle w=\frac{x-y}{|x-y|}$. Then

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S |u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^{|x-y|}|\nabla u(x+tw)|dtdy$

$\displaystyle \leq \frac{1}{|S|}\int_\Omega\int_0^{\infty}|\nabla u(x+tw)|\chi_{\{t<|x-y|\}}dtdy$

$\displaystyle \leq \frac{1}{|S|} \int_0^\infty\int_{|w|=1}\int_0^d|\nabla u(x+tw)|\chi_{\{t<|x-rw|\}}r^{n-1}dwdrdt$

$\displaystyle \leq \frac{d^n}{n|S|}\int_0^\infty\int_{|w|=1}|\nabla u(x+tw)|\chi_{\{t<|x-\xi(w)|\}}dwdt$

$\displaystyle =\frac{d^n}{n|S|}\int_\Omega \frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$

where $\xi(w)$ is the intersection of ray $x+tw$ with $\partial \Omega$ (it is unique because $\Omega$ is convex).

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### Weak (1,1) boundedness of Hardy Littlewood maximal function

$f$ is a locally integrable function in $\mathbb{R}^d$, define the maximal function as

$\displaystyle M(f)=\sup\limits_{x\in B}\frac{1}{|B|}\int_B|f(y)|dy$

here $B$ is an  arbitrary ball.

$\textbf{Thm: }$ Maximal operator is weakly (1,1) bounded. That is $\exists C=C(d)>0$ such that for $\forall\lambda>0$,

$\displaystyle |\{x:M(f)(x)>\lambda\}|\leq\frac{C||f||_1}{\lambda}$

$\textbf{Proof: }$ Let $E_\lambda=\{x:M(f)(x)>\lambda\}$, $K\subset E_\lambda$ is a compact set.

Then $K\subset\bigcup\limits_{i=1}^N B_i$, where $B_i=B_i(x_i)$ with $\displaystyle \frac{1}{|B_i|}\int_{B_i}|f(y)|dy>\lambda$. Apply the Vitali covering lemma

$\textbf{Vitali covering lemma: }$ Given $\{B_1,\cdots,B_n\}$ a finite collection of balls in $\mathbb{R}^n$, then there exists a subcollection $B_{ij}$, $j=1,\cdots,s$ disjoint from each other and

$\bigcup\limits_{i=1}^N B_i\subset \bigcup\limits_{j=1}^s 3B_{i_j}$

We can conclude that  $K\subset \bigcup\limits_{j=1}^s 3B_{i_j}$. So

$\displaystyle |K|\leq \sum\limits_{j=1}^s 3^d|B_{i_j}|\leq \frac{3^d}{\lambda}\sum\limits_{j=1}^s\int_{B_{i_j}}|f(y)|dy\leq \frac{3^d||f||_1}{\lambda}$

Since $K$ is arbitrary, we know $|E_\lambda|$ is bounded by the right hand side.$\hfill\square$

Actually the decay of maximal function is a little faster than $\frac{1}{\lambda}$

$\textbf{Thm: }$ There exists a constant $C=C(d)>0$ such that

$\displaystyle |\{x:M(f)(x)>\lambda\}|\leq \frac{C}{\lambda}\int_{|f|>\frac{\lambda}{2}}|f(y)|dy$

$\textbf{Proof: }$ Define

$\displaystyle g=\begin{cases}\frac{\lambda}{2},\quad |f(x)|>\frac{\lambda}{2}\\ 0, \quad |f(x)|\leq\frac{\lambda}{2}\end{cases}$

Then $|f(x)|\leq |g(x)|+\frac{\lambda}{2}$. This means

$\displaystyle \{x:M(f)(x)>\lambda\}\subset \{x:M(g)(x)>\frac{\lambda}{2}\}$

So $\displaystyle |\{x:M(f)(x)>\lambda\}|\leq |\{x:M(g)(x)>\frac{\lambda}{2}\}|\leq \frac{C}{\lambda}\int_{|f|>\frac{\lambda}{2}}|f(y)|dy$

$\hfill\square$

### Boundedness of fractional integration operator

Suppose $f$ is locally integrable, define the fractional integration operator

$\displaystyle I_\alpha(f)(x)=\int_{\mathbb{R}^n}\frac{f(y)}{|x-y|^{n-\alpha}}dy$, $0<\alpha

$\mathbf{Thm:}$ Suppose $f\in L^p(\mathbb{R}^n)$ and $0<\alpha, then

$\displaystyle ||I_\alpha(f)||_q\leq C(p,n,\alpha)||f||_p$

where $\displaystyle \frac{1}{q}=\frac{1}{p}-\frac{\alpha}{n}$ and $p>1$. If $p=1$, then $I_\alpha(f)$ is a weak type $(1,\frac{n}{n-\alpha})$.

$\bf{Lemma:}$ Suppose $\psi:\mathbb{R}^n\to\mathbb{R}^n$ is radially decreasing function, then

$\displaystyle \left|\int_{\mathbb{R}^n}f(x)\psi(x-y)dy\right|\leq M(f)(x)||\psi||_1$

when $||\psi||_1<\infty$.

$\textbf{Proof of Theorem:}$ Write

$\displaystyle I_\alpha(f)(x)=\int_{\mathbb{R}^n}\frac{f(x-y)}{|y|^{n-\alpha}}dy=\int_{|y|\leq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy+\int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy$

By the lemma,

$\displaystyle \int_{|y|\leq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq M(f)(x)\int_{|y|\leq A}\frac{1}{|y|^{n-\alpha}}dy\leq \frac{nw_n}{\alpha}A^\alpha M(f)(x)$

By holder inequality, if $p>1$

$\displaystyle \int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq ||f||_p\left(\int_{|y|\geq A}\frac{1}{|y|^{(n-\alpha)p'}}\right)^\frac{1}{p'}\leq \left(\frac{qw_n}{p'}\right)^\frac{1}{p'}A^{-\frac{n}{q}}||f||_p$

If $p=1$

$\displaystyle \int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq A^{-\frac{n}{q}}||f||_1$ with $\displaystyle q=\frac{n}{n-\alpha}$.

Combing the preceeding result,

$\displaystyle |I_\alpha(f)(x)|\leq C(n,p,\alpha)(A^{-\frac{n}{q}}||f||_p+A^\alpha M(f)(x))$ for $\forall\,A>0$

Choose $A=Mf(x)^{-\frac{p}{n}}||f||_p^{\frac{p}{n}}$,

$\displaystyle |I_\alpha(f)(x)|\leq C(n,\alpha,p)M(f)(x)^{\frac{p}{q}}||f||_p^{1-\frac{p}{q}}$

When $p>1$,  maximal function is bounded from $L^p$ to $L^p$,

$\displaystyle ||I_\alpha(f)||_q\leq C(n,\alpha,p)||f||_p^\frac{p}{q}||f||_p^{1-\frac{p}{q}}=C(n,\alpha,p)||f||_p$

When $p=1$, maximal function is weak (1,1),

$\displaystyle |\{x||I_\alpha(f)(x)|>\lambda|\}|\leq \{x|C(n,\alpha)M(f)(x)^{\frac{n-\alpha}{n}}||f||_1^{\frac{\alpha}{n}}>\lambda\}$

$\displaystyle =|\{x|M(f)(x)>\left(\frac{\lambda}{C(n,\alpha)||f||_1^{\frac{\alpha}{n}}}\right)^{\frac{n}{n-\alpha}}\}|$

$\displaystyle \leq 3^n||f||_1\left(\frac {C(n,\alpha)||f||_1^{\frac{\alpha}{n}}}{\lambda} \right)^\frac{n}{n-\alpha}=C'(n,\alpha)\left(\frac{||f||_1}{\lambda}\right)^\frac{n}{n-\alpha}$

$\square$

When the domain is not $\mathbb{R}^n$ but a bounded one, we can obtain the boundedness of $\displaystyle \frac{1}{q}<\frac{1}{p}-\frac{\alpha}{n}$.

$\textbf{Thm: }$ Suppose $f\in L^p(\Omega)$, $p>1$. We have $I_\alpha(f)$ is bounded from $L^p$ to $L^q$  with

$\displaystyle 0\leq \delta=\frac{1}{p}-\frac{1}{q}<\frac{\alpha}{n}$, $1\leq q\leq \infty$

$\displaystyle ||I_\alpha f||_q\leq Cw_n^{1-\alpha/n}|\Omega|^{\alpha/n-\delta}||f||_p$

$\textbf{Remark:}$ Loukas Grafakos: Modern fourier analysis.

### Mixed norm and basic property

In the analysis of evolution equation, we will constantly treat this kind of function $f(x,t)$ where $x\in \mathbb{R}^d$, $t\in \mathbb{R}$. Considering the difference of time and space variable, we will introduce this kind of norm

$\displaystyle ||f||_{L^q_t L^r_x}=\left(\int_\mathbb{R}\left(\int_{\mathbb{R}^d}|f|^rdx\right)^{\frac{q}{r}}dt\right)^{\frac 1q}$  for $q,r\geq 1$

This kind of norm also has similar properties of $L^p$ norm.

$\mathbf{Proposition 1:}$ $\displaystyle \iint |fg|dxdt \leq ||f||_{L^q_t L^r_x}||g||_{L^{q'}_t L^{r'}_x}$ here $q$ and $q'$ are conjugate  and so are $r,r'$.

$\mathbf{Proof:}$ $\displaystyle \iint |fg|dxdt =\int\left(\int |fg|dx\right)dt\leq \int\left(||f||_{L^r_x}||g||_{L^{r'}_x}\right)dt\leq||f||_{L^q_t L^r_x}||g||_{L^{q'}_t L^{r'}_x}$

$\mathbf{Corollary 1:}$ $\displaystyle ||f||_{L^q_t L^r_x}\leq ||f||_{L^{q_1}_t L^{r_1}_x}||f||_{L^{q_2}_t L^{r_2}_x}$, where $\displaystyle \frac 1q=\frac {1}{q_1}+\frac{1}{q_2}$ and $\displaystyle \frac 1r=\frac{1}{r_1}+\frac{1}{r_2}$. All the powers are bigger than or equal 1.

$\mathbf{Proposition 2:}$ $\displaystyle ||f||_{L^q_t L^r_x}=\sup_{||g||_{L^{q'}_t L^{r'}_x}}\iint fgdxdt$

$\mathbf{Proof:}$ From the proposition 1, LHS$\leq ||f||_{L^q_t L^r_x}$.

Define $\displaystyle g=sgn(f)|f|^{r-1}\left(\int |f|^r\right)^{\frac{q}{r}-1}||f||^{\frac{1}{q}-1}_{L^q_t L^r_x}$ when $||f||_{L^q_t L^r_x}\neq 0$, otherwise the proposition holds trivially.
Then $||g||_{L^{q'}_t L^{r'}_x}=1$ and $\displaystyle \iint fgdxdt=||f||_{L^q_t L^r_x}$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### Various proofs of Hilbert transform: Hardy’s inequality, Minkowski integral inequality

$\mathbf{Problem:}$ For $x,y>0$, define

$\displaystyle Tf(x)=\int_0^\infty \frac{f(y)}{x+y}dy$

prove that $T:L^p\to L^p$ is a bounded operator.

Proof by Hardy’s inequality

Hardy’s inequality: Let $1\leq p<\infty$ and $f\geq 0$

1. $\forall\, \alpha<-1$

$\displaystyle \int_0^\infty\left(\int_0^x f(t)dt\right)^px^\alpha dx\leq \left(\frac{p}{1+|\alpha|}\right)^p\int_0^\infty f^p(t)t^{\alpha+p}dt$

2. $\forall\, \alpha>-1$

$\displaystyle \int_0^\infty\left(\int_x^\infty f(t)dt\right)^px^\alpha dx\leq \left(\frac{p}{1+\alpha}\right)^p\int_0^\infty f^p(t)t^{\alpha+p}dt$

$\mathbf{Proof:}$ Split $Tf(x)$ as $\displaystyle Tf(x)=\int_0^x \frac{f(y)}{x+y}dy+\int_{x}^\infty\frac{f(y)}{x+y}dy$. Then

$\displaystyle \left\|\int_0^x \frac{f(y)}{x+y}dy\right\|_p^p=\int_0^\infty \left|\int_0^x \frac{f(y)}{x+y}dy\right|^pdx\leq \int_0^\infty \left(\frac{1}{x}\int_0^x |f(y)|dy\right)^pdx\leq C_p\int_0^\infty|f(y)|^pdy$

Here we used Hardy’s inequality case (1) with $\alpha=-p<-1$.

$\displaystyle \left\|\int_x^\infty \frac{f(y)}{x+y}dy\right\|_p^p=\int_0^\infty \left|\int_x^\infty \frac{f(y)}{x+y}dy\right|^pdx\leq \int_0^\infty \left(\int_x^\infty\frac{|f(y)|}{y}dy\right)^pdx\leq C_p\int_0^\infty|f(y)|^pdy$

Here we used Hardy’s inequality case (2) with $\alpha=0$.

So $\displaystyle \|Tf\|_p\leq \left\|\int_0^x\frac{f(y)}{x+y}dy\right\|_p+\left\|\int_x^\infty\frac{f(y)}{x+y}dy\right\|_p\leq C\|f\|_p$.

$\text{Q.E.D}\hfill \square$

Proof by Minkowski integral inequality

$\mathbf{Proof:}$ Changing variable by $y=tx$

$\displaystyle \int_0^\infty \frac{f(y)}{x+y}dy=\int_0^\infty\frac{f(tx)}{1+t}dt$

So by Minkowski Integral Inequality

$\displaystyle \|Tf\|_p=\left( \int_0^\infty \left(\int_0^\infty \frac{f(tx)}{1+t}dt\right)^pdx\right)^\frac{1}{p}\leq \int_0^\infty\left(\int_0^\infty \left|\frac{f(tx)}{1+t}\right|^pdx\right)^\frac{1}{p}dt\\=\int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}\left(\int_0^\infty|f(y)|^p\right)^\frac{1}{p}dt=\|f\|_p\int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}dt$

Since $p>1$, $\displaystyle \int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}dt<\infty$. We are done.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

Pay attention to the tricks using here. The proof by Hardy’s inequality is hinted from Prof. Chanillo’s notes. For Minkowski method, see stein’s book.