Let $f(x) \in C^2[0,1]$, with $f(0)=f(1)=0$, and $f(x)\neq 0$ when $x \in(0,1)$. Show that $\int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \geq 4.$
Prove that $\int_0^1 \left| \frac{f^{''}(x)}{f(x)} \right| dx \geq4$.
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0And, of course, trying to prove it is a lot easier than actually proving it:) – 2012-03-10
4 Answers
Here's another answer, which avoids reducing to the case in which $f$ is concave. It is plainly enough to prove that \sup_{x \in [0,1]}{|f(x)|} \leq \frac{1}{4}I,\quad \text{where} \quad I := \int_0^1|f''(x)|\,dx. \tag{1} First of all, since $f(0) = f(1) = 0$ and $f$ is nonzero in $(0,1)$, we know that the supremum on the left-hand side is attained at some $c \in (0,1)$, and moreover, that f'(c) = 0. By using Taylor's theorem with remainder (which is really just repeated integration by parts) to expand $f$ around the point $c$, we have f(x) = f(c) + f'(c)(x-c) + \int_c^x (x - t)f''(t)\,dt = f(c) + \int_0^c (x - t)f''(t)\,dt, for any $x \in [0,1]$. Successively taking $x = 0$ and $x = 1$ gives f(c) = -\int_0^c tf''(t)\,dt = -\int_c^1 (1-t)f''(t)\,dt, because $f(0) = f(1) = 0$. This means that |f(c)| \leq c\int_0^c |f''(t)|\,dt \quad \text{and} \quad |f(c)| \leq (1-c) \int_c^1 |f''(t)|\,dt. \tag{2} Since \int_0^c|f''(t)|\,dt + \int_c^1|f''(t)|\,dt = I = (1-c)I + cI, we must have either \int_0^c|f''(t)|\,dt \leq (1-c) I or \int_c^1|f''(t)|\,dt \leq c I. Either way $(2)$ shows that $ |f(c)| \leq c(1-c)I = \frac{1}{4}I - \left(\frac{1}{2} - c\right)^2I \leq \frac{1}{4}I, $ and $(1)$ is therefore proved.
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0@NickStrehlke you are my hero. – 2012-03-19
Without loss of generality, assume that $f(x)\ge0$ on $[0,1]$. Furthermore, we can assume that f''(x)\le0. If not, we can replace $f$ by $g$ where the graph of $g$ is the convex hull of the graph of $f$. Note that where $g(x)\not=f(x)$, g''(x)=0, therefore, \int_0^1\left|\frac{g''(x)}{g(x)}\right|\,\mathrm{d}x\le\int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x.
Suppose that f'(0)=a and f'(1)=-b. Since $f$ is concave, $f(x)\le ax$ and $f(x)\le b(1-x)$. Therefore, $ \max_{[0,1]}f(x)\le\frac{ab}{a+b}\tag{1} $ Furthermore, \begin{align} \int_0^1|f''(x)|\,\mathrm{d}x &\ge\left|\int_0^1f''(x)\,\mathrm{d}x\right|\\[6pt] &=|f'(1)-f'(0)|\\[6pt] &=a+b\tag{2} \end{align} Therefore, since $\min\limits_{\mathbb{R^+}}\frac{(1+t)^2}{t}=4$, \begin{align} \int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x &\ge\frac{1}{\max\limits_{[0,1]}f(x)}\int_0^1|f''(x)|\,\mathrm{d}x\\ &\ge\frac{(a+b)^2}{ab}\\ &=\frac{(1+b/a)^2}{b/a}\\ &\ge4\tag{3} \end{align}
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0@Sam: $g$ is $C^1$ and piecewise $C^2$, and this proof works for that class of functions, which includes $C^2$. If we consider a sawtooth (or triangle) function as the limit of such functions, then the integral of $\frac{f''}{f}$ around a vertex is the jump in $f'$ divided by $f$ at that vertex. Thus, if we include that in $\int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x$ the result holds for those functions, too. – 2012-03-10
There is an elementary proof for this inequation, by mean-value theorem.
CASE I: $f(x)\geq 0 $ on $[0,1]$
$f(0)=f(1)=0$ There is a point $x_0\in (0,1)$ s.t $\max_{[0,1]}f(x)=f(x_0)$
By mean-value theorem ,there are $\lambda_1 \in (0,x_0),\lambda_2 \in (x_0,1)$
s.t $\begin{align} f(x_0)-f(0)=\int_0^{x_0} f'(x) dx = (x_0 - 0)f'(\lambda_1)\Rightarrow f'(\lambda_1)=\frac{f(x_0)}{x_0}\end{align}$
$\begin{align} f(1)-f(x_0)=\int_{x_0}^{1} f'(x) dx = (1-x_0)f'(\lambda_2)\Rightarrow f'(\lambda_2)=\frac{-f(x_0)}{1-x_0}\end{align}$
$\begin{align} \int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \ge \int_{\lambda_1}^{\lambda_2} \left| \frac{f''(x)}{f(x)} \right| dx \ge \int_{\lambda_1}^{\lambda_2} \frac{|f''(x)|}{f(x_0)} dx \\= \frac {1}{f(x_0)} \int_{\lambda_1}^{\lambda_2} |f''(x)| \ge \frac {1}{f(x_0)} |\int_{\lambda_1}^{\lambda_2} f''(x) |\end{align}$
And $\begin{align} \frac {1}{f(x_0)}| \int_{\lambda_1}^{\lambda_2} f''(x)| =\frac {1}{f(x_0)}|f'(\lambda_2)-f'(\lambda_1)| =\frac {1}{f(x_0)}|\frac{-f(x_0)}{1-x_0}-\frac{f(x_0)}{x_0}|=\frac {1}{(1-x_0)(x_0)}\ge 4\end{align}$
CASE II: $f(x)$ may be nagetive on $[0,1]$.
Since $\forall x\in [0,1],|f(x)|\le M$ for some M,let $g(x) = f(x) + 2M$.
Then $\forall x\in [0,1],|g(x)|\ge M \ge |f(x)|$ and $g''(x)=f''(x),g(x)\ge 0$ on $[0,1]$
$\begin{align} \int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx =\int_0^1 \left| \frac{g''(x)}{f(x)} \right| dx\ge \int_0^1 \left| \frac{g''(x)}{g(x)} \right| dx \ge 4\end{align} $ (By CASE I)
Let $f(c)=\max_{x\in[0,1]}f(x)$. Then By Mean Value Theorem, $f(c)-f(0)=f'(a)c, \quad f(1)-f(c)=f'(b)(1-c)$ so we have $f'(a)=\frac{f(c)}{c},\quad f'(b)=-\frac{f(c)}{1-c}$. Then $\int^1_0\frac{|f"(x)|}{|f(x)|}dx\ge\frac{1}{|f(c)|}\int^1_0|f''(x)|dx\ge\frac{1}{|f(c)|}\int^b_a|f''(x)|dx\ge\frac{1}{|f(c)|}|\int^b_af''(x)dx|=\frac{1}{|f(c)|}|f'(b)-f'(a)|=\frac{1}{|f(c)|}|\frac{f(c)}{c}+\frac{f(c)}{1-c}|=\frac{1}{c(1-c)}\ge\frac{1}{4}$