# Unique solution on subspaces whose union is dense implies unique solution globally?

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.

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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 $-4$.