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.