$\int_a^b |f(x)||g(x)| dx \leq \left(\int_a^b |f(x)|^p dx\right)^{\frac1p}\left(\int_a^b |g(x)|^q dx\right)^{\frac{1}{q}}$

Question

Let $p\gt 1,q\gt 1$ be the dual indices, $\frac1p + \frac1q = 1$ and let $X$ be the space of all continuous functions on $[a,b]$ with two real numbers $a\lt b$. $f(x)$ and $g(x)$ are continuous functions on $[a,b]$

I want to prove that:

$$\int_a^b |f(x)||g(x)| dx \leq \left(\int_a^b |f(x)|^p dx\right)^{\frac1p}\left(\int_a^b |g(x)|^q dx\right)^{\frac{1}{q}}$$

I have been suggested to use young's inequality, but I can't find the relevant thing to use. It seems on wiki to be very relevant to use 'Young's inequality for convolutions', but I have never dealt with any $L^p$ spaces, should I just learn these, or is this not what they were referring?

You need the first inequality in this article http://en.wikipedia.org/wiki/Young%27s_inequality — Crostul, Apr 07 '15 at 10:22
Apply Young's inequality $ab \leq \frac{a^p}{p} + \frac{b^q}{q}$ for well-chosen $a$ and $b$. — , Apr 07 '15 at 10:22
@V.C. I suppose so, but I have never dealt with measure theory — Understand, Apr 07 '15 at 10:23
@maxerize There are two forms, and I don't think the one in your comment is the correct one — Understand, Apr 07 '15 at 10:24
@Understand I assure you that this one works, see Jack's answer below — , Apr 07 '15 at 11:32

Jack D'Aurizio · Accepted Answer · 2015-04-08T10:35:14.013

Do you remember how we can prove Cauchy-Schwarz through interpolation?

Since $|xy|\leq\frac{x^2+y^2}{2}$ we have: $$ \| f\cdot g\|_1 \leq \frac{\|f\|_2^2}{2}+\frac{\|g\|_2^2}{2} \tag{1} $$ but the LHS is just the same if we replace $f$ with $\lambda f$ and $g$ with $\frac{1}{\lambda}g$, so: $$ \| f\cdot g\|_1 \leq \frac{\lambda^2\|f\|_2^2}{2}+\frac{\|g\|_2^2}{2\lambda^2} \tag{2} $$ and by choosing $\lambda$ in such a way the two terms in the RHS of $(2)$ are equal, i.e. $\lambda=\sqrt{\frac{\|g\|_2}{\|f\|_2}}$, we get: $$ \| f\cdot g\|_1 \leq \|f\|_2 \cdot \|g\|_2 \tag{3}$$ that is the usual Cauchy-Schwarz inequality. If we start with the Young inequality: $$ |xy|\leq \frac{|x|^p}{p}+\frac{|y|^q}{q}\tag{4} $$ and follow exactly the same interpolation steps, we end with: $$ \| f\cdot g\|_1 \leq \|f\|_p\cdot \|g\|_q \tag{5} $$ that is the wanted Holder's inequality.

Thank you, this is a very good answer! – Understand Apr 08 '15 at 05:19 — Understand, Apr 08 '15 at 05:19

$\int_a^b |f(x)||g(x)| dx \leq \left(\int_a^b |f(x)|^p dx\right)^{\frac1p}\left(\int_a^b |g(x)|^q dx\right)^{\frac{1}{q}}$

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