Proof of Riesz-Fischer Theorem using the completeness of $L^1$ to infer completeness of $L^p$.

Question

One can infer the completeness of $L^p$ by using the homeomorphism $\varphi: L^p \to L^1$ defined by $f \mapsto |f|^{p-1} f$. This rests on the inequality: $$ 2^{1-p} || f- g||_p^p \leq ||\varphi(f) -\varphi(g)||_1\leq p (||f||_p + ||g||_p)^{p-1} ||f-g ||_p$$ Which I am not sure how to prove, does anyone know where this inequality comes from or have any hints on how to show this? Thanks.

Answer 1

For the first inequality:

First we prove that $u^{p}-1-2^{1-p}(u-1)^{p}\geq 0$ for $u\geq 1$.

Let $\phi(u)=u^{p}-1-2^{1-p}(u-1)^{p}$. For $u>1$, then $\phi(u)=\phi(u)-\phi(1)=p(\xi^{p-1}-2^{1-p}(\xi-1)^{p-1})$, where $\xi\in(1,u)$. Note that \begin{align*} \xi^{p-1}&> 2^{1-p}(\xi-1)^{p-1}\\ \xi&> 2^{-1}(\xi-1)\\ 2\xi&>\xi-1\\ \xi&>-1 \end{align*} each step is valid if and only if, so $\xi^{p-1}-2^{1-p}(\xi-1)^{p-1}>0$ and hence $\phi(u)>0$ for $u>1$, and hence $\phi(u)\geq 0$ for $u\geq 1$.

Now let $a\geq b>0$, then $a/b\geq 1$ and hence \begin{align*} 2^{1-p}\left(\dfrac{a}{b}-1\right)^{p}\leq\left(\dfrac{a}{b}\right)^{p}-1, \end{align*} so \begin{align*} 2^{1-p}(a-b)^{p}\leq a^{p}-b^{p}, \end{align*} this inequality also holds for $b=0$, so we conclude that $2^{1-p}(a-b)^{p}\leq a^{p}-b^{p}$ for $a\geq b\geq 0$.

Assume now that $a\geq 0\geq b$, by Jensen's inequality, \begin{align*} \left(\dfrac{a+(-b)}{2}\right)^{p}\leq\dfrac{1}{2}a^{p}+\dfrac{1}{2}(-b)^{p}, \end{align*} we conclude that \begin{align*} 2^{1-p}(a-b)^{p}\leq a^{p}+(-b)^{p}=\bigg|a|a|^{p-1}-b|b|^{p-1}\bigg|. \end{align*} We conclude the general inequality that \begin{align*} 2^{1-p}|f-g|^{p}\leq\bigg|f|f|^{p-1}-g|g|^{p-1}\bigg|, \end{align*} taking integrals both sides, we have \begin{align*} 2^{1-p}\|f-g\|_{p}^{p}&=2^{1-p}\int|f-g|^{p}\\ &\leq\int\bigg|f|f|^{p-1}-g|g|^{p-1}\bigg|\\ &=\|\varphi(f)-\varphi(g)\|_{1}. \end{align*}

For the second inequality:

For $a,b\geq 0$, then $|a^{p}-b^{p}|=p|a-b|\xi^{p-1}\leq p|a-b|(a+b)^{p-1}$ for some $\xi$ lies in between $a,b$, this is by Mean Value Theorem.

Assume first that $f,g\geq 0$, then \begin{align*} |f^{p}-g^{p}|\leq p|f-g|(f+g)^{p-1}, \end{align*} then by Holder's inequality, we have \begin{align*} \int|f^{p}-g^{p}|&\leq p\left(\int|f-g|^{p}\right)^{1/p}\left(\int((f+g)^{p-1})^{p/(p-1)}\right)^{(p-1)/p}\\ &=p\|f-g\|_{p}\|f+g\|_{p}^{p-1}\\ &\leq p\|f-g\|_{p}(\|f\|_{p}+\|g\|_{p})^{p-1}, \end{align*} where the last inequality is by Minkowski's inequality.

Now consider the general $f,g$. If $x$ is such that $f(x),g(x)\geq 0$, then \begin{align*} ||f(x)|^{p-1}f(x)-|g(x)|^{p-1}g(x)|&=|f(x)^{p}-g(x)^{p}|\leq p|f(x)-g(x)|(|f(x)|+|g(x)|)^{p-1}, \end{align*} if $x$ is such that $f(x)\geq 0\geq g(x)$, then \begin{align*} ||f(x)|^{p-1}f(x)-|g(x)|^{p-1}g(x)|&=f(x)^{p}+(-g(x))^{p}\\ &\leq(f(x)-g(x))^{p}\\ &\leq p(f(x)-g(x))(f(x)-g(x))^{p-1}\\ &=p|f(x)-g(x)|(|f(x)|+|g(x)|)^{p-1}, \end{align*} so in either case, we have the inequality that \begin{align*} ||f|^{p-1}f-|g|^{p-1}g|\leq p|f-g|(|f|+|g|)^{p-1}, \end{align*} the rest reasoning is similar to the first one by taking Holder's inequality and then Minkowski's inequality.