7

... is the result in $L_{p}$?

A remark in my notes says yes but I can't see how to verify it.

As was pointed out to me in a previous question I asked last night, I need to show that the following integral is finite:

$$\int_{-\pi}^{\pi}|\int_{-\pi}^{\pi}f(t-s)\phi_{n}(s)ds|^{p}dt < \infty$$.

One of the properties of a summability kernel is that there exists a $C > 0$ such that $\int_{-\pi}^{\pi}|\phi_{n}(t)|dt\leq C$ for every $n\geq 1$. I feel like this could help if I could get $\phi$ by itself

roo
  • 5,800

1 Answers1

18

The way I usually prove Young's Inequality is using a couple of applications of Hölder's inequality to prove $$ \left|\int f(x)\;g(x)\;h(x)\;\mathrm{d}x\right|\le\|f\|_u\|g\|_v\|h\|_w\tag{1} $$ where $\frac1u+\frac1v+\frac1w=1$. Then apply $(1)$ in a tricky way to show $$ \left|\iint f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\le\|f\|_p\|g\|_q\|h\|_r\tag{2} $$ where $\frac1p+\frac1q+\frac1r=2$. Taking the supremum of inequality $(2)$ over all $h\in L^r$ such that $\|h\|_r=1$ says that $\|f\ast g\|_s\le\|f\|_p\|g\|_q$ where $\frac{1}{r}+\frac{1}{s}=1$, that is $\frac{1}{s}=\frac{1}{p}+\frac{1}{q}-1$.

Tricky Application of $\mathbf{(1)}$: Since $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$, we have the following 7 relations: $$ \begin{align} \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= 1\\ \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{p}\text{ so that }p\left(1-\frac{1}{r}\right) + p\left(1-\frac{1}{q}\right) = 1\\ \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) &= \frac{1}{q}\text{ so that }q\left(1-\frac{1}{r}\right) + q\left(1-\frac{1}{p}\right) = 1\\ \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{r}\text{ so that }r\left(1-\frac{1}{p}\right) + r\left(1-\frac{1}{q}\right) = 1 \end{align} $$ Therefore, $$ \begin{align} &\left|\iint f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\\ &\le\iint|f(x-y)|\;|g(y)|\;|h(x)|\;\mathrm{d}y\;\mathrm{d}x\\ &={\small\iint\underbrace{|f(x-y)|^{p(1-1/r)}|g(y)|^{q(1-1/r)}}_{\large\text{in }L^w\text{ where }\frac1r+\frac1w=1}\;\underbrace{|g(y)|^{q(1-1/p)}|h(x)|^{r(1-1/p)}}_{\large\text{in }L^u\text{ where }\frac1p+\frac1u=1}\;\underbrace{|f(x-y)|^{p(1-1/q)}|h(x)|^{r(1-1/q)}}_{\large\text{in }L^v\text{ where }\frac1q+\frac1v=1}\;\mathrm{d}y\;\mathrm{d}x}\\ &\le\left(\iint|f(x-y)|^p|g(y)|^q\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/r}\\ &\times\left(\iint|g(y)|^q|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/p}\\ &\times\left(\iint|f(x-y)|^p|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/q}\\ &=\left(\int|f(x)|^p\;\mathrm{d}x\right)^{1/p}\left(\int|g(x)|^q\;\mathrm{d}x\right)^{1/q}\left(\int|h(x)|^r\;\mathrm{d}x\right)^{1/r} \end{align} $$
robjohn
  • 353,833
  • ♦ sorry could you explain the last passage you have done? Why is this true? I really don't get this. I've tried to interchange the variables thanks to fubini, but the last integral ((∫∫|f(x−y)|p|h(x)|rdydx)) depends only to x... – Bman72 Jan 20 '14 at 13:55
  • 1
    @Ale: Use Fubini to see that $$\begin{align}\int\int|f(x-y)|^p|g(y)|^q,\mathrm{d}y,\mathrm{d}x &=|f|_p^p\int|g(y)|^q,\mathrm{d}y\ &=|f|_p^p|g|_q^q\end{align}$$ and similarly for the other integrals. – robjohn Jan 20 '14 at 14:44
  • The inequality (1) is false; if it were true, then for functions $f,g,h$ which are non zero and $fgh$ is non zero, applying the inequality to $f(\lambda x), g(\lambda x),h(\lambda x)$ for $\lambda \in \mathbb{R}$, you get $\lambda^{-n}|fgh|_1 \leq \lambda^{-n(1/p + 1/q + 1/r)}|f|_p|g|_q|h|_r$, which can't be true for every $\lambda$ unless $1/p + 1/q + 1/r = 1$. – Matt Rigby Sep 25 '14 at 15:36
  • 1
    @MattRigby: you are correct. A simple way to see this is to use $h(x)=1$ ($r=\infty$) and note that it gives the wrong version of Hölder. This argument works, but somewhere I copied something wrong. I will fix this. – robjohn Sep 25 '14 at 19:03
  • 1
    @MattRigby: there are two triples of exponents. In the Hölder case, $\frac1u+\frac1v+\frac1w=1$. In the Young case, $\frac1p+\frac1q+\frac1r=2$. I have separated the exponents to keep things less confusing. Thanks for catching that. – robjohn Sep 25 '14 at 19:57
  • No worries, looks good now :) – Matt Rigby Sep 25 '14 at 21:52