On the proof of the Hölder inequality.

Question

In my attempts of finding different proofs for the Hölder inequality, I have found none like this one that I have derived for myself using the Young inequality: if $p\in]1;\infty[$ and $\frac{1}{p}+\frac{1}{q}=1$, then $\forall(x,y)\in(\mathbb{R}^{+*})^2$: \begin{equation} \frac{x^p}{p}+\frac{y^q}{q}\geq xy \end{equation} Naturally I would like to know whether my proof is indeed valid. The proof goes as follows: let $f,g:[a,b]\to\mathbb{R}$ such that $a<b$. We can state with confidence that: $$|f(t)g(t)|=(\lambda |f(t)|)(\frac{1}{\lambda}|g(t)|)\leq \frac{\lambda^p}{p}|f(t)|^p+\frac{1}{q\lambda^q}|g(t)|^q$$ $\forall\lambda>0$. Integrating on both sides and applying the tirangle inequality for integrals leaves us with: $$\big{|}{\int_a^b f(t)g(t)\ dt\big{|}}\leq \frac{\lambda^p}{p}\int_a^b|f(t)|\ dt+\frac{1}{q\lambda^q}\int_a^b|g(t)|^q\ dt$$ Now consider the function: $$\psi(\lambda)=\frac{\lambda^p}{p}\int_a^b|f(t)|\ dt+\frac{1}{q\lambda^q}\int_a^b|g(t)|^q\ dt$$ And let us denote $F=\int_a^b|f(t)|\ dt$ and $G=\int_a^b|g(t)|^q\ dt$ for the sake of shorthand. Taking the derivative yields: $$\psi'(\lambda)=\lambda^{p-1}F-\lambda^{-q-1}G$$ Solving for $\psi'(\lambda)=0$ shows that $\psi$ has only one minimum on $\mathbb{R}^{+*}$: $$\psi'(\lambda)=0\Leftrightarrow \lambda^{p-1}F=\lambda^{-q-1}G$$ $$\Leftrightarrow \lambda=\left(\frac{G}{F}\right)^{\frac{1}{p+q}}$$ Plugging this value into $\psi$ gives us its minimum, which yields: $$\psi(\left(\frac{G}{F}\right)^{\frac{1}{p+q}})=\frac{F}{p}\left(\frac{G}{F}\right)^{\frac{p}{p+q}}+\frac{G}{q}\left(\frac{G}{F}\right)^{\frac{-q}{p+q}}$$ Factorising by $G\left(\frac{G}{F}\right)^{\frac{-q}{p+q}}$ leaves us with: $$\psi(\left(\frac{G}{F}\right)^{\frac{1}{p+q}})=G\left(\frac{G}{F}\right)^{\frac{-q}{p+q}}$$ Now note how: $$\frac{q}{p+q}=\frac{\frac{p}{p-1}}{p+\frac{p}{p-1}}=\frac{1}{p}\wedge\frac{p}{p+q}=\frac{1}{q}$$ This leaves us with: $$\psi(\left(\frac{G}{F}\right)^{\frac{1}{p+q}})=G^{\frac{1}{q}}F^{\frac{1}{p}}\geq\big{|}{\int_a^b f(t)g(t)\ dt\big{|}}$$ Which concludes the proof for Hölder's inequality: $$\big{|}{\int_a^b f(t)g(t)\ dt\big{|}}\leq \left(\int_a^b|f(t)|^p\ dt\right)^{\frac{1}{p}}\left(\int_a^b|g(t)|^q\ dt\right)^{\frac{1}{q}}$$ If this is valid it's the simplest proof that I can find using elementary calculus. Thank you for your time!

Answer 1

Your overall argument seems correct. I leave you a few comments: you should verify that your critical point is indeed a minimum point; in a few places the exponent $p$ is missing in $|f(x)|^p$; and, finally, your "wedge notation" is, I guess, not conventionan (you were probably going for '&' or something?). Other than that, seems nice proof!

I have seen a short proof before (which is very classical, not mine) that also uses only Young's inequality to prove Hölder. Denote, like you did, $$F := \int_{a}^{b} |f(t)|^{p} \, \mathrm{d} t \quad\text{ and }\quad G := \int_{a}^{b} |g(t)|^{q} \, \mathrm{d} t.$$ The inequality is trivial if either $f \equiv 0$ or $g\equiv 0$. So, it remains to check the case in which both are nonzero functions. Assume first that $F=G=1$. Then, by Young, $$ \int_{a}^{b} f(t)g(t) \, \mathrm{d} t \le \dfrac{1}{p}\int_{a}^{b} |f(t)|^{p} \, \mathrm{d} t + \dfrac{1}{q} \int_{a}^{b} |g(t)|^{q} \, \mathrm{d} t = \dfrac{1}{p} + \dfrac{1}{q} = 1 = F\cdot G. $$ Now the general case follows by considering $\tilde{f} = F^{-1/p} f$ and $\tilde{g} = G^{-1/q} g$, since these functions satisfy $$\int_{a}^{b} |\tilde{f}(t)|^{p} \, \mathrm{d} t = \frac{1}{F} \int_{a}^{b} |f(t)|^{p} \, \mathrm{d} t = 1$$ $$\int_{a}^{b} |\tilde{g}(t)|^{p} \, \mathrm{d} t = \frac{1}{G} \int_{a}^{b} |g(t)|^{p} \, \mathrm{d} t = 1$$ We conclude $$\int_{a}^{b} \tilde{f}(t)\tilde{g}(t) \, \mathrm{d} t \le 1$$ which implies $$\int_{a}^{b} {f}(t){g}(t) \, \mathrm{d} t \le F \cdot G.$$

Answer 2

Simplifying the proof of @Diego Marcon (and fixing an error), let: %$$ F := \int_{a}^{b} |f(t)|^{p} \, \mathrm{d} t \quad\quad G := \int_{a}^{b} |g(t)|^{q} \, \mathrm{d} t, $$ and (≈normalised) $\quad \tilde{f} := F^{-1/p}\!f$, $\quad \tilde{g} := G^{-1/q}\!g % \quad\implies\quad % \int_{a}^{b} |\tilde{f}(t)|^{p} \, \mathrm{d} t \ =\, \int_{a}^{b} |\tilde{g}(t)|^{q} \, \mathrm{d} t \ =\, 1$.

The inequality is trivial if $f \!=\! 0$ or $g \!=\! 0$, so assume $f$, $g\,\ne\,0$ and bounded. By Young (*), $$ \tfrac{1}{F^{1/p}\,G^{1/q}}\!\int_{a}^{b}\!\! |f(t)g(t)| \,\mathrm{d}t \ =\, \int_{a}^{b}\!\! |\tilde{f}(t)\tilde{g}(t)| \, \mathrm{d}t \ \overset{(*)}{\le}\, \tfrac{1}{p}\!\int_{a}^{b}\!\! |\tilde{f}(t)|^{p} \, \mathrm{d} t + \tfrac{1}{q} \!\int_{a}^{b}\!\! |\tilde{g}(t)|^{q} \, \mathrm{d}t \ =\, \tfrac{1}{p} \!+\! \tfrac{1}{q} \ =\, 1$$ Hence $$\int_{a}^{b}\! |f(t)g(t)| \, \mathrm{d} t \ \leq\ F^{1/p} G^{1/q} \ =\ \big(\int_{a}^{b}\! |f(t)|^{p}\big)^{1/p} \big(\int_{a}^{b}\! |g(t)|^{q}\big)^{1/q}\tag*{$\square$}$$