# Why is n-th Fréchet derivative symmetric?

Let $V,W$ be nonzero normed spaces over $\mathbb{K}$.

Let $E$ be open in $V$ and $f:E\rightarrow W$ be a twice Fréchet-differentiable function.

Then, $D^2 f: E\rightarrow \mathscr{L}_2(V^2,W)$ is symmetric. That is, at any point $p\in E$, $((D^2 f)(p) )(x,y)= ((D^2 f) (p))(y,x)$.

It is not that I don't understand the proof, but I don't understand why it must hold. What's the geometric meaning of higher order Fréchet derivatives?

First order Fréchet derivative $(D f)(p)$ is a function that approximates $f(x)$ where $x$ near $p$, by means of a linear transformation. From the definition $(D^2 f)(p)$ is a linear approximation of $(Df)(x)$ where $x$ is near $p$. However, I'm not sure what does $D^2 f$ say about $f$.

What does $D^n f$ represent of $f$ exactly?

• You may think $f(p) + Df(p)x + \frac{1}{2} D^2 f(p)(x,x)$ as the (best) quadratic approximation to $f(p+x)$. – user251257 Aug 26 '15 at 0:22
• On a slightly lower level, understanding why $\partial^2 f/\partial x\partial y =\partial^2 f/\partial y\partial x$ for a $C^2$ function is a thorny question. It's hard to "see" geometrically. It is an analytic fact. We do come back to the approximating quadratic function, but still ... – Ted Shifrin Aug 26 '15 at 1:30