Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ denote the space of all $f : [0,1] \to {\mathbb R}$ such that the second derivative $f''$ is continuous except on a finite set, equipped with the norm $N(f)=|f(0)|+|f’(0)|+||f''||_{\infty}$ (where $||f''||_{\infty}$ is $||f''||_{\infty}={\sf sup}(|f''(x)|, x \in [0,1])$, the supremum norm). Let $W$ be the “affine subspace” of $V$ defined by the inequations $f(0)=f’(0)=0$ and $f(1)=1, f’(1)=0$. The problem is to find the smallest possible value $M$ for $||f''||_{\infty}$ when $f\in W$,and to describe the functions that attain this bound (if they exist).

I can show that $ M =4$, see below. And it would seem from my proof that the optimal solution I found is unique, but all my attempts to make this explicit failed. I’m also curious to know whether it is true that if $f''$ is continuous everywhere and $||f''||_{\infty}$ is near to the optimum $4$, then $f$ must be near my optimal solution (in terms of the $N$ norm defined above).

My optimal solution is as follows : we take $f''$ to be $+4$ on $[0,\frac{1}{2}]$ and $-4$ on $[\frac{1}{2},1]$, so that

$$ f(x)=\begin{cases} 2x^2 & \text{if} x\in[0,\frac{1}{2}] \\ -1+4x-2x^2 & \text{if} x\in[\frac{1}{2},1] \\ \end{cases} $$

then we have $||f''||_{\infty}=4$. So $M \leq 4$. Conversely, we must show $||f''||_{\infty} \geq 4$ for any $f\in W$. Let $L_n$ be the subspace of functions in $V$, such that $f''$ is constant on each interval $I_k=[\frac{k}{n},\frac{k+1}{n}]$. Since the union of all the $L_n \cap W$ is dense in $W$, we may assume that $f\in L_n \cap W$. Denote by $a_k$ the value of $f''$ on $I_k$. Integrating and summing on $k$, we then have

$$ f'(1)-f’(0)=\frac{1}{n}\Bigg(\sum_{k=1}^{n}a_k\Bigg) \ \text{and} \ f(1)-f(0)=\frac{1}{2n^2}\Bigg(\sum_{k=1}^{n}(2n-2k+1)a_k\Bigg), $$

and hence

$$ (2n)(f(1)-f(0))- (n-1)(f'(1)-f’(0))=\frac{1}{n}\Bigg(\sum_{k=1}^{n}(n-2k+2)a_k\Bigg), $$

The left-hand side is equal to $2n$ and the absolute value of the right-hand side is bounded by $\frac{||f''||_{\infty}}{n}\sum_{k=1}^{n}|n-2k+2|$. So

$$ ||f''||_{\infty} \geq \frac{2n^2}{\sum_{k=1}^{n}|n-2k+2|} $$

For even $n$, the sum in the denominator evaluates to $\frac{n^2}{2}$, and hence $||f''||_{\infty} \geq \frac{2n^2}{\frac{n^2}{2}}=4$, as wished.

share|cite|improve this question
up vote 1 down vote accepted

You can simply say that for $f\in V$ there is a finite set $0=a_0<a_1<\ldots < a_n=1$ such that $f''$ is exists an is continuous on each $(a_k,a_{k+1})$. From the MVT $f'(t)-f'(a_k)=f''(\xi)(t-a_k)$ for $a_k<t<a_{k+1}$, hence the assumption of $||f''||_\infty\le 4$ leads to $|f'(t)-f'(a_k)|\le 4|t-a_k|$ and by induction on $k$: $f'(a_k)\le 4a_k$ and thus $f'(x)\le 4x$ just as if we were allowed to apply the MVT to all of $[0,1]$. Similarly, $f'(x)\ge 4(1-x)$ from the right end. Therefore $1-2(1-x)^2\le f(x)\le 2x^2$. Especially, $f(\frac12)=\frac12$. On $[0,\frac12]$ consider $g(x)=f(x)-2x^2$. Then $g(x)\le0$ and $g'(x)\le 0$ for $0\le x \le \frac12$. From $-g(x)=g(\frac12)-g(x)=g'(\xi)(\frac12-x)\le0$ we conclude that $g(x)=0$ on $[0,\frac12]$, A similar argument works for $[\frac12,1]$ hence the optimal function $$f_0(x)=\begin{cases}2x^2&0\le x \le\frac12\\1-2(1-x)^2&\frac12\le x\le 1\end{cases} $$ you found is indeed unique.

I am not sure about your second question for the restriction of $N$ to $W$ is simply $||f''||_\infty$. For $a\in\mathbb R$ Consider $$g_a(x)=\begin{cases}2x^2&0\le x\le \frac12\\ a\left(x-\frac12\right)^3+2x^2&x\ge \frac12 \end{cases}$$ Then $g_a$ is $C^2$ and $g_a''(x)=4$ for $0\le x\le \frac12$ and $g_a''(x)=6a\left(x-\frac12\right)+4$ for $x\ge\frac12$. For $0<h<\frac12$, let $a=-\frac2{3h}$. Then $0\le g_a''(x)\le 4$ for $0\le x\le \frac 12+h$ and $g_a''(\frac12+h)=0$. With this choice of $a$ let $$f_h(x)=\begin{cases} \frac{g_a((1+2h)x)}{2g_a(\frac12+h)} & 0\le x\le\frac12\\ \frac{g_a((1+2h)(1-x))}{2g_a(\frac12+h)} & \frac12\le x\le1 \end{cases}$$ Then $f_h$ is $C^2$. Also, $2g_a(\frac12+h)=-\frac43 h^2+4(\frac12+h)^2\to1$ as $h\to 0$, hence $||f_h||\to 4$ as $h\to 0$. Thus we see that there are $C^2$ functions in $W$ that have $N(f)$ arbitrarly close to $N(f_0)$. However, if $f''$ is continuous, then necessarily $||f-f_0||_\infty\ge 4$ because $f_0''$ jumps from $+4$ to $-4$.

share|cite|improve this answer
Thank you Hagen, you answered everything. I guess you forgot to put the second derivatives inside the norm on the last line. – Ewan Delanoy Oct 3 '12 at 12:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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