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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.

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    $\begingroup$ This might work: Let $h > 0$. replace $f$ by $hf$, $g$ by $g/h$. Now minimize the resulting inequality over $h > 0$. $\endgroup$
    – Mason
    Sep 12, 2022 at 19:28
  • $\begingroup$ so you want to say that I have to consider $(hf)^{\alpha}(g/h)^{1-\alpha} \leq ahf + bg/h$? $\endgroup$
    – vitalmath
    Sep 12, 2022 at 19:36
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    $\begingroup$ integrate that then optimize. Actually, you want the $h$s to cancel on the left, so maybe this isn't the way to go $\endgroup$
    – Mason
    Sep 12, 2022 at 19:38
  • $\begingroup$ Hmm ok, I see... $\endgroup$
    – vitalmath
    Sep 12, 2022 at 19:59
  • $\begingroup$ 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". $\endgroup$ Sep 12, 2022 at 23:05

2 Answers 2

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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: \begin{equation} \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} \end{equation}

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.

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This is somewhat using a sledgehammer to crack a nut, but you can use something called the "weighted AM-GM inequality" and the result drops out.

$$ {1\over 1-\alpha+\alpha} \left( \alpha\left(\int f\right)^\alpha+ (1-\alpha) \left(\int g\right)^{1-\alpha}\right)\geq \sqrt[1-\alpha+\alpha]{ \left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha} } $$ The $1-\alpha+\alpha$ in the above is just $1$, and both the division and the root are redundant, so $$ \alpha \left( \int f\right)^\alpha+ (1-\alpha) \left(\int g\right)^{1-\alpha}\geq \left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha} $$

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