1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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
3
$\begingroup$

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 $g(s,t) = H(k,\,h) \cdot s \cdot t.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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