Let $E$ and $F$ be normed spaces. What can you say of a function $f:A\subseteq E\to F$ with $A$ open in $E$ twice differentiable, if $D^2 f$ is constant?

This is a very open question that do not know how to answer ... help me?

  • $\begingroup$ What are $E$ and $F$? $\endgroup$ Jul 5 '13 at 16:15
  • $\begingroup$ Normed Spaces only.. @JonasMeyer $\endgroup$ Jul 5 '13 at 16:16
  • $\begingroup$ the OP asks for $D^2f$ to be constant, not $D^1f$. I canceled my answer. Even if $E=\mathbb R^2$ and $F=\mathbb R$ some computations have to be done... $\endgroup$
    – Avitus
    Jul 5 '13 at 16:24
  • $\begingroup$ I guess the derivative is in this sense? $\endgroup$
    – newbie
    Jul 5 '13 at 19:41

I assume that $A$ is connected. Otherwise, the below would apply on each connected component of $A$ without necessarily holding globally. Furthermore, I assume that differentiable means that the Fréchet derivative exists, and $D^kf$ refers to the Fréchet derivatives.

The answer is then as one would expect it to be, for any $a \in A$, we have

$$f(x) = f(a) + Df\lvert_a(x-a) + \frac12 D^2f\lvert_a(x-a,x-a)$$

for all $x \in A$, where we have identified $L(E,\, L(E,\,F)) \cong L^2(E,\,F)$ the space of continuous linear mappings from $E$ to $L(E,\,F)$ with the space $L^2(E,\,F)$ of continuous bilinear mappings from $E\times E \to F$, as usual.

First, for a differentiable $g \colon A \to G$, where $G$ is any normed space, with constant derivative $Dg \equiv H \in L(E,\,G)$, we show that

$$g(x) = g(a) + H(x-a),\quad x \in A$$

for all $a \in A$.

Fix an arbitrary $a \in A$ and suppose that $\lVert x - a\rVert < \operatorname{dist}(a,\,\complement A)$. Let $\varphi \colon [0,\,1] \to A$ be given by $\varphi(t) = a + t\cdot (x-a)$.

For every $\lambda \in G^\ast$, consider the function $h \colon [0,\,1] \to \mathbb{K}$ given by $h = \lambda \circ g \circ \varphi$.

$$h'(t) = D\lambda\lvert_{g(\varphi(t))} \Bigl( Dg\lvert_{\varphi(t)} \bigl(D\varphi\lvert_t (\frac{\partial}{\partial t})\bigr)\Bigr) = \lambda \bigl(H(x-a)\bigr)$$

is independent of $t$, hence $h(1) = h(0) + h'(0)$, or $\lambda\bigl(g(x)\bigr) = \lambda\bigl(g(a)\bigr) + \lambda\bigl(H(x-a)\bigr)$.

Since $G^\ast$ separates points on $G$, that means $g(x) = g(a) + H(x-a)$.

We have proved that on each ball contained in $A$, now let us make that global.

For any fixed $a \in A$, let

$$M_a := \{ x \in A \colon g(x) = g(a) + H(x-a)\}.$$

By continuity, $M_a$ is closed in $A$, and by the above, $a \in \overset{\circ}{M_a}$.

Now let $\xi \in M_a$. In $M_\xi$, we have

$$g(x) = g(\xi) + H(x - \xi) = \bigl(g(a) + H(\xi - a)\bigr) + H(x-\xi) = g(a) + H(x-a)$$

by linearity, hence $M_\xi \subset M_a$. But $M_\xi$ is a neighbourhood of $\xi$, hence $M_a$ is also open (in $A$). $A$ is connected, therefore $M_a = A$.

Apply the above to $Df$, to obtain

$$Df\lvert_x = Df\lvert_a + H(x-a)$$

for all $a,\, x \in A$.

Again, fix an arbitrary $a \in A$ and consider $x$ with $\lVert x-a\rVert < \operatorname{dist}(a,\,\complement A)$. Let $\varphi$ as above, and for each $\lambda \in F^\ast$, consider $\lambda \circ f \circ \varphi$.

$$\begin{align}(\lambda \circ f \circ \varphi)'(t) &= \lambda \Bigl(Df\lvert_{\varphi(t)}\bigl(\varphi'(t)\bigr)\Bigr) = \lambda \Bigl(\bigl(Df\lvert_a + H(\varphi(t)-a)\bigr)(x-a)\Bigr)\\ &= \lambda\bigl(Df\lvert_a (x-a) + t\cdot H(x-a,\,x-a)\bigr)\\ &= \lambda \bigl(Df\lvert_a(x-a)\bigr) + t\cdot\lambda\bigl(H(x-a,\,x-a)\bigr). \end{align}$$

Hence $\lambda\bigl(f(x) - f(a) - Df\lvert_a(x-a) - \frac12 H(x-a,\,x-a)\bigr) = 0$, for all $\lambda \in F^\ast$, hence

$$f(x) = f(a) + Df\lvert_a(x-a) + \frac12 H(x-a,\,x-a)$$

in a neighbourhood of $a$.

Again, set $M_a := \{ x \in A\colon f(x) = f(a) + Df\lvert_a(x-a) + \frac12 H(x-a,\,x-a) \}$. By continuity, $M_a$ is closed in $A$, an analogous argument as above shows that $M_a$ is open in $A$, hence $M_a = A$.

For $\xi \in M_a$ and $x \in M_\xi$, we have

$$\begin{align}f(x) &= f(\xi) + Df\lvert_\xi (x-\xi) + \frac12 H(x-\xi,\,x-\xi)\\ &= f(a) + Df\lvert_a (\xi-a) + \frac12 H(\xi-a,\,\xi-a) + Df\lvert_a(x-\xi) + H(\xi-a,\,x-\xi) + \frac12 H(x-\xi,\,x-\xi)\\ &= f(a) + Df\lvert_a(x-a) + \frac12 H(\xi-a,\,x-a) + \frac12 H(x-a,\,x-\xi)\\ &= f(a) + Df\lvert_a(x-a) + \frac12 H(x-a,\,x-a). \end{align}$$

One thing needs to be made explicit, the argument to show the openness here uses the fact that $D^2f$ is symmetric. That need not be the case if differentiability is not Fréchet differentiability, so if differentiability is meant in some other sense, the result might not hold.

As a post script, the symmetry of the second Fréchet derivative can be seen by considering, for $h,\, k \in E$, (and assuming without loss of generality $0 \in A$) the function

$$\begin{align} g(s,t) &= f(sh + tk) - f(sh) - f(tk) + f(0)\\ &= \int_0^t Df\lvert_{sh + \tau k}(k) - Df\lvert_{\tau k}(k)\, d\tau\\ &= \int_0^t \biggl(\int_0^s D^2f\lvert_{\tau k + \sigma h}(h)\, d\sigma\biggr)(k)\, d\tau\\ &= H(h,\,k)\cdot s\cdot t. \end{align}$$

(The integrals here are Pettis integrals.)

Grouping the other way, $\bigl(f(sh+tk) - f(tk)\bigr) - \bigl(f(sh) - f(0)\bigr)$, leads to $g(s,t) = H(k,\,h) \cdot s \cdot t.$


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