# Deduce the Hölder Inequality

If $$f, g$$ are two functions $$\geq 0$$ in an interval $$I$$ such that the integrals $$\int_I f(t) d t$$ and $$\int_I g(t) d t$$ are convergent, the integral $$\int_I(f(t))^\alpha(g(t))^{1-\alpha} d t$$ is convergent and we have $$\int_I(f(t))^\alpha(g(t))^{1-\alpha} d t \leq a \int_I f(t) d t+b \int_I g(t) d t .$$ Deduce the Hölder inequality $$\int_I(f(t))^\alpha(g(t))^{1-\alpha} d t \leq\left(\int_I f(t) d t\right)^\alpha\left(\int_I g(t) d t\right)^{1-\alpha}$$

My attempt: By Young Inequality we know that for $$0 \leq \alpha \leq 1$$, we have $$x^{\alpha} y^{1- \alpha} \leq ax + by,$$where $$a + b = 1$$. Thus if we set $$x = f(t)$$ and $$y=g(t)$$, then we get $$f(t)^{\alpha}g(t)^{1-\alpha} \leq af(t) + bg(t).$$ If we integrate over $$I$$, we get the wanted inequality. I don't see how to proceed to get the Hölder inequality now.

• This might work: Let $h > 0$. replace $f$ by $hf$, $g$ by $g/h$. Now minimize the resulting inequality over $h > 0$. Sep 12, 2022 at 19:28
• so you want to say that I have to consider $(hf)^{\alpha}(g/h)^{1-\alpha} \leq ahf + bg/h$? Sep 12, 2022 at 19:36
• integrate that then optimize. Actually, you want the $h$s to cancel on the left, so maybe this isn't the way to go Sep 12, 2022 at 19:38
• Hmm ok, I see... Sep 12, 2022 at 19:59
• There is a Terry Tao blog post on this. He calls it an "amplification" of an inequality, using symmetry. The blog post is called "Amplification, arbitrage and the tensor power trick". Sep 12, 2022 at 23:05

This is maybe easier to see, if we define $$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$$ $$\norm{f}_p = \left( \int_{I} |f(t)|^{p} \, dt \right)^{1/p}$$ and $$p, q$$ such that $$\alpha = 1/p$$ and $$1-\alpha = 1/q$$.
Furthermore, by replacing $$f$$ by $$f^{p}$$ and $$g$$ by $$g^{q}$$, the inequality can be rewritten in simpler form.
Finally, by multiplying by suitable constants, we may assume that $$\norm{f}_p = 1$$ and $$\norm{g}_q = 1$$ and deduce that: $$$$\int_{I} f(t) g(t) \, dt \leq a \norm{f}_p^{p} + b \norm{g}_q^{q} = 1 = \norm{f}_p \norm{g}_q \tag{1}$$$$
Remark. I basically just rewrote everything into a more standard form using norms and these substitutions. From which Hölder's inequality clearly follows. Replacing $$f^{p}$$ by $$f$$ and $$g^{q}$$ by $$g$$ and substituting back $$\alpha$$ in (1) we get Hölder's inequality in the form it was written in the question.
$${1\over 1-\alpha+\alpha} \left( \alpha\left(\int f\right)^\alpha+ (1-\alpha) \left(\int g\right)^{1-\alpha}\right)\geq \sqrt[1-\alpha+\alpha]{ \left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha} }$$ The $$1-\alpha+\alpha$$ in the above is just $$1$$, and both the division and the root are redundant, so $$\alpha \left( \int f\right)^\alpha+ (1-\alpha) \left(\int g\right)^{1-\alpha}\geq \left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha}$$