Proving $\int_0^1\frac{\vert f(x)\rvert^2}{x^2}\,\mathrm dx\le4\int_0^1{\vert f'(x)\rvert^2}\,\mathrm dx$ when $f\in\mathcal C^1([0,1])$ and $f(0)=0.$

Let us assume $f \in \mathcal{C}^1([0,1])$ and $f(0)=0.$ Prove that $$\int_{0}^{1} \frac{\vert f(x) \rvert^2}{x^2}dx \le 4 \int_{0}^{1} {\vert f'(x) \rvert^2}dx.$$

By integrating by parts I obtained the following

$$\int_{0}^{1} \frac{\vert f(x) \rvert^2}{x^2}dx = -\frac{1}{x} \lvert f(x) \rvert^2\Big|_0^1+2\int_{0}^{1} \frac{f(x)|f'(x)|}{x |f(x)|} \le 2\int_{0}^{1} \frac{|f'(x)|}{x } $$

but I'm not sure of the result and its usefulness, it's easy Calculus, but I can't go on. Any suggestions?

Solution 1:

We have that $$ \int_0^1 \frac{f^2(x)}{x^2}dx =\int_0^1\frac{1}{x^2}\left(\int_0^x2f(t)f'(t)dt\right) dx\\ =2\int_0^1f(t)f'(t)\left(\int_t^1\frac{dx}{x^2} dx\right) dt\\ =2\int_0^1f'(t)\cdot \left(\frac{f(t)}{t}(1-t)\right)dt. $$ Then by Cauchy-Schwarz inequality, $$\left(\int_0^1f'(t)\cdot \left(\frac{f(t)}{t}(1-t)\right)dt\right)^2 \leq \int_0^1(f'(t))^2 dt \cdot \int_0^1 \frac{f^2(t)}{t^2}(1-t)^2dt\\ \leq \int_0^1(f'(t))^2 dt \cdot \int_0^1 \frac{f^2(t)}{t^2}dt,$$ where we used also the fact that $0\leq (1-t)^2\leq 1$ for $t\in[0,1]$. Hence $$\left(\int_0^1 \frac{f^2(x)}{x^2}dx\right)^2\leq 4\int_0^1(f'(t))^2 dt \cdot \int_0^1 \frac{f^2(t)}{t^2}dt$$ which implies that $$\int_0^1 \frac{f^2(x)}{x^2}dx\leq 4\int_0^1(f'(t))^2 dt.$$

P.S. The costant $4$ is optimal: take $f(x)=x^{1/2+1/n}$ then $$\int_0^1 \frac{f^2(x)}{x^2}dx=\frac{n}{2} \quad\mbox{and}\quad\int_0^1(f'(x))^2 dx=\frac{(n+2)^2}{8n}.$$ Therefore $$\lim_{n\to+\infty}\frac{\frac{n}{2}}{\frac{(n+2)^2}{8n}}=4.$$

Solution 2:

You have almost done it with your integration by parts attempt, just missing one simple last step. Again as you started was \begin{align} I&=\int_0^1\frac{f(x)^2}{x^2}\,dx=\underbrace{-\frac{f(x)^2}{x}\Bigg|_0^1}_{\le 0}+2\int_0^1\frac{f(x)}{x}f'(x)\,dx\le 2\int_0^1\frac{f(x)}{x}f'(x)\,dx\le\\ &\le 2\left(\int_0^1\frac{f(x)^2}{x^2}\,dx\right)^{1/2}\left(\int_0^1 f'(x)^2\,dx\right)^{1/2}=2\sqrt{I}\left(\int_0^1 f'(x)^2\,dx\right)^{1/2}. \end{align} Now divide by $\sqrt{I}$ and square all.

Solution 3:

Similar to Robert Z's answer, but with a bit of simplification $$ \begin{align} \int_0^1\frac{f(x)^2}{x^2}\,\mathrm{d}x &=\int_0^1\frac2{x^2}\int_0^xf(t)\,f'(t)\,\mathrm{d}t\,\mathrm{d}x\tag{1}\\ &=\int_0^1f(t)\,f'(t)\int_t^1\frac2{x^2}\,\mathrm{d}x\,\mathrm{d}t\tag{2}\\ &=2\int_0^1f(t)\,f'(t)\left(\frac1t-1\right)\mathrm{d}t\tag{3}\\ &\le2\int_0^1|f(t)|\,|f'(t)|\,\frac1t\,\mathrm{d}t\tag{4}\\ &\le2\left[\int_0^1\frac{f(t)^2}{t^2}\,\mathrm{d}t\int_0^1f'(t)^2\,\mathrm{d}t\right]^{1/2}\tag{5} \end{align} $$ Explanation:
$(1)$: $f(x)^2=f(x)^2-f(0)^2=2\int_0^xf(t)\,f'(t)\,\mathrm{d}t$
$(2)$: change order of integration
$(3)$: integrate in $x$
$(4)$: $0\le\frac1t-1\le\frac1t$ on $[0,1]$
$(5)$: Cauchy-Schwarz

Divide $(5)$ by $\left(\int_0^1\frac{f(x)^2}{x^2}\,\mathrm{d}x\right)^{1/2}$ and square to get $$ \int_0^1\frac{f(x)^2}{x^2}\,\mathrm{d}x\le4\int_0^1f'(t)^2\,\mathrm{d}t\tag{6} $$

Variational Approach and Proof of Sharpness

Maximize $$ \int_0^1\frac{f(x)^2}{x^2}\,\mathrm{d}x\tag{7} $$ under the constraint that $$ \int_0^1f'(x)^2\,\mathrm{d}x=1\tag{8} $$ That is, find the $f$ so that $$ 0=\int_0^1\frac{f(x)\,\delta f(x)}{x^2}\,\mathrm{d}x\tag{9} $$ for every $\delta f(x)$ so that $$ \begin{align} 0 &=\int_0^1f'(x)\,\delta f'(x)\,\mathrm{d}x\\ &=-\int_0^1f''(x)\,\delta f(x)\,\mathrm{d}x\tag{10} \end{align} $$ which requires that there is a $\lambda$ so that $f''(x)=\lambda\frac{f(x)}{x^2}$ which is satisfied by $x^\alpha$. Plugging in $f(x)=\frac{\sqrt{2\alpha-1}}\alpha x^\alpha$ gives $$ \begin{align} \int_0^1f'(x)^2\,\mathrm{d}x &=\frac{2\alpha-1}{\alpha^2}\int_0^1\alpha^2x^{2\alpha-2}\,\mathrm{d}x\\ &=\frac{2\alpha-1}{\alpha^2}\frac{\alpha^2}{2\alpha-1}\\[6pt] &=1\tag{11} \end{align} $$ and $$ \begin{align} \int_0^1\frac{f(x)^2}{x^2}\,\mathrm{d}x &=\frac{2\alpha-1}{\alpha^2}\int_0^1x^{2\alpha-2}\,\mathrm{d}x\\ &=\frac{2\alpha-1}{\alpha^2}\frac1{2\alpha-1}\\ &=\frac1{\alpha^2}\tag{12} \end{align} $$ Thus, the critical functions, $f(x)=\frac{\sqrt{2\alpha-1}}{\alpha}x^\alpha$, have a constant of $\frac1{\alpha^2}$. This seems to indicate that we can't bound $\int_0^1\frac{f(x)^2}{x^2}\,\mathrm{d}x$ by any multiple of $\int_0^1f'(x)^2\,\mathrm{d}x$ until we notice that for the integrals to converge, we need $\alpha\gt\frac12$. This gives the constant of $4$ as in the question. In fact, this shows that $4$ cannot be improved.