Let $f$ be a $C^2$ function in $[0,1]$ such that $f(0)=f(1)=0$ and $f(x)\neq 0\,\forall x\in(0,1).$ Prove that

$$\int_0^1 \left|\frac{f{''}(x)}{f(x)}\right|dx\ge4$$

  • 1
    $\begingroup$ I think you mean $f(0) = f(1) = 0$, not $f(x) = f(1) = 0$. $\endgroup$ – Michael Albanese Oct 14 '12 at 5:03
  • $\begingroup$ Yep. f(0)=f(1)=0. Sorry about the typo. $\endgroup$ – Christmas Bunny Oct 14 '12 at 5:21
  • $\begingroup$ I can show that, if $f\in C^2\[0,1\]$ and $f'(0)=f'(1)=0$ then exist a point $t\in \[0,1\]$ such that, $f''(t)>4|f(1)-f(0)|$ $\endgroup$ – Salech Rubenstein Oct 14 '12 at 5:34
  • $\begingroup$ This is a duplicate of a question that has been asked before but I am unable to find the other question. $\endgroup$ – user17762 Oct 14 '12 at 5:56

Just some considerations. Without loss of generality, we can assume that $f(x)$ is positive on $(0,1)$. Let $\operatorname{graph}(g)$ be the convex hull of $\operatorname{graph}(f)$. We have $g(x)\in C^2([0,1])$ and

$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|g''(x)|}{g(x)}dx,$$

so we can also assume that $f$ is a concave function over $[0,1]$, with $f'(0)=\alpha>0,f'(1)=-\beta<0$ and a unique maximum in $x_0\in(0,1)$, for which $f'(x_0)=0$ and $f(x_0)=1$. Let now $\operatorname{graph}(h)$ be the envelope of the tangent lines to $\operatorname{graph}(f)$ for $x\in\left\{0,x_0,1\right\}$. For any $\epsilon>0$ there is a function $u\in C^2([0,1])$ such that $|u-h|<\epsilon$ and:

$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|u''(x)|}{u(x)}dx=\alpha+\beta-O(\epsilon),$$

but $\alpha+\beta$ must be greater than $4$, since $\max u$ is below the $y$-coordinate of the intersection of the tangent lines in $x=0$ and $x=1$, so

$$ 1 = \max u \leq \frac{\alpha\beta}{\alpha+\beta} \leq_{AM-GM} \frac{\alpha+\beta}{4}.$$


Proving the Lower Bound

Without loss of generality, assume that $f(x)\gt0$ for $x\in(0,1)$.

Suppose that $f(x_0)=y_0=\max\limits_{x\in[0,1]}f(x)$. Then, $f'(x_0)=0$.

By the Mean Value Theorem, for some $x_1\in(0,x_0)$, $f'(x_1)=\frac{y_0}{x_0}$. Therefore, $$ \int_0^{x_0}|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{x_0} $$ Furthermore, for some $x_2\in(x_0,1)$, $f'(x_2)=-\frac{y_0}{1-x_0}$. Therefore, $$ \int_{x_0}^1|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{1-x_0} $$ Since $f(x)\le y_0$, $$ \begin{align} \int_0^1\left|\,\frac{f''(x)}{f(x)}\,\right|\,\mathrm{d}x &\ge\frac{\frac{y_0}{x_0}+\frac{y_0}{1-x_0}}{y_0}\\ &=\frac1{x_0}+\frac1{1-x_0}\\ &=\frac1{\frac14-\left(x_0-\frac12\right)^2}\\[3pt] &\ge4 \end{align} $$

The Lower Bound is Sharp

Let $$ f_a(x)=\sin^{-1}\left(\frac{\sin(\pi x)}{1+a^2\sin^2(\pi x)}\right) $$ enter image description here

then $$ \lim_{a\to0}\int_0^1\left|\,\frac{f_a''(x)}{f_a(x)}\,\right|\,\mathrm{d}x=4 $$ since $f_a''(x)$ is tends to $0$ except near $x=\frac12$, and $\int_0^1f_a''(x)\,\mathrm{d}x$ tends to $-2\pi$, whereas $f_a\!\left(\frac12\right)$ tends to $\frac\pi2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.