Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

... 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

share|improve this question
Yes, apply Young's inequality with $r=p$ and $q = 1$, see also: en.wikipedia.org/wiki/Convolution#Integrable_functions –  t.b. Oct 5 '11 at 0:21
Thank you this is very helpful. –  Kyle Schlitt Oct 5 '11 at 0:28

1 Answer 1

up vote 10 down vote accepted

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 $$ \left|\int\int f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\le\|f\|_p\|g\|_q\|h\|_r\tag{2} $$ 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, $$ \begin{align} &\left|\int\int f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\\ &\le\int\int|f(x-y)|\;|g(y)|\;|h(x)|\;\mathrm{d}y\;\mathrm{d}x\\ &={\small\int\int\underbrace{|f(x-y)|^{p(1-1/r)}|g(y)|^{q(1-1/r)}}_{\large\text{in }L^w\text{ where }\frac1r+\frac1w=1}\;\underbrace{|g(y)|^{q(1-1/p)}|h(x)|^{r(1-1/p)}}_{\large\text{in }L^u\text{ where }\frac1p+\frac1u=1}\;\underbrace{|f(x-y)|^{p(1-1/q)}|h(x)|^{r(1-1/q)}}_{\large\text{in }L^v\text{ where }\frac1q+\frac1v=1}\;\mathrm{d}y\;\mathrm{d}x}\\ &\le\left(\int\int|f(x-y)|^p|g(y)|^q\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/r}\\ &\times\left(\int\int|g(y)|^q|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/p}\\ &\times\left(\int\int|f(x-y)|^p|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/q}\\ &=\left(\int|f(x)|^p\;\mathrm{d}x\right)^{1/p}\left(\int|g(x)|^q\;\mathrm{d}x\right)^{1/q}\left(\int|h(x)|^r\;\mathrm{d}x\right)^{1/r} \end{align} $$

share|improve this answer
♦ 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... –  Ale Jan 20 at 13:55
@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 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 at 15:36
@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 at 19:03
@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 at 19:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.