Showing if $f' \in L^2([1,\infty),m)$, then $\frac{f(x)}{x} \in L^2([1,\infty),m)$

Question

Let $f:[1,\infty) \to \Bbb{R}$ be such that $f(1)=0$, $f'$ exists, is bounded, and $f' \in L^2([1,\infty),m)$, then if $g(x)=\frac{f(x)}{x}$, we have $g \in L^2([1,\infty),m)$.

Furthermore, show this fails when we replace $L^2$ with $L^1$.

Attempt

For the first part, I was planning on using integration by parts then realized it gets too messy with the absolute values and all so I tried a change of variables, would I be correct in putting say $x=yt$ then $dx=dt$.

And I get stuck at this step.

For the second part, I found a function and I want to know if it works. I found $f$ to being

$$f(x)=\frac{\ln x}{x}.$$

Clearly $f(1)=0$. And $f'=\frac{1-\ln x}{x^2}$ and this has integral from $1$ to $\infty$ of $0$ while $f(x)/x=\ln x$ diverges. So our $g$ wouldnt be in $L^1([1,\infty),m)$. Shoot n evermind $f(x)/x=\frac{\ln x}{x^2}$..I just noticed that sorry so neither one of my parts for this problem are ok..wow I was so excited

Answer 1

Hardy’s inequality.

We have for each $R>1$, and $\epsilon>0$, \begin{align} \int_{1}^{R}\frac{|f(x)|^2}{x^2}\,dx&=\left[-\frac{|f(x)|^2}{x}\right]_{1}^{R}-\int_1^R\left(\frac{-1}{x}\right)2ff’\,dx\\ &=-\frac{|f(R)|^2}{R}+2\int_1^R\frac{1}{x}ff’\,dx\\ &\leq 0+\int_1^R\epsilon\frac{|f(x)|^2}{x^2}+\frac{|f’(x)|^2}{\epsilon}\,dx. \end{align} Here, we have used the trivial inequality $2|ab|\leq |a|^2+|b|^2$ with $a= \sqrt{\epsilon}\frac{f(x)}{x}$ and $b=\frac{f’(x)}{\sqrt{\epsilon}}$. So, by subtracting the $\int_1^R\epsilon\frac{|f(x)|^2}{x^2}\,dx$ term from both sides, we can rearrange the inequality (once $\epsilon$ is small enough, say $0<\epsilon<1$) and then take the limit $R\to\infty$ to get \begin{align} \int_1^{\infty}\frac{|f(x)|^2}{x^2}\,dx&\leq\frac{1}{\epsilon(1-\epsilon)}\int_1^{\infty}|f’(x)|^2\,dx. \end{align} If you really wanted to, you could now optimize over $\epsilon$ as well to get a better constant.

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So, the point is that when doing the integration by parts, the boundary term at $x=1$ vanishes by hypothesis, while the one at $\infty$ appears with the correct sign so that the inequality points in the right way at the end. So, the upshot is that the boundary terms don’t matter. The rest is the standard trick of absorbing part of the RHS into a good LHS.

Answer 2

This deals with the case $f'\in L_2(1,\infty)$ with $f(1)=0$. An application of the generalized Minkowski inequality yields \begin{align} \Big(\int^\infty_1\Big|\frac{1}{x}\int^\infty_0 \mathbb{1}_{[1,x)}(t)\,f'(t)\,dt\Big|^2\,dx\Big)^{1/2}&= \Big(\int^\infty_1\Big|\frac{1}{x}\int^\infty_0 \mathbb{1}_{[1/x,1)}(t/x)\,f'(xt/x)\,dt\Big|^2\,dx\Big)^{1/2}\\\ &=\Big(\int^\infty_1\Big|\int^\infty_0 \mathbb{1}_{[1/x,1)}(u)\,f'(xu)\,du\Big|^2\,dx\Big)^{1/2}\\ &\leq\int^\infty_0\Big(\int^\infty_1|f'(xu)|^2\mathbb{1}_{[1/x,1)}(u)\,dx\Big)^{1/2}\,du\\ &\leq\int^\infty_0\Big(\int^\infty_0|f'(xu)|^2\mathbb{1}(xu>1)\mathbb{1}(u<1)\,dx\Big)^{1/2}\,du\\ &\stackrel{s=xu}{=}\int^1_0u^{-1/2}\Big(\int^\infty_1|f'(s)|^2\,ds\Big)^{1/2}\,du\\ &\leq2\|f'\|_{L_2([1,\infty)} \end{align}