# Symmetry of the second derivative

For the purposes of this question, a function $f$ is differentiable at $x\in \mathbb{R}^d$ iff (i) the directional derivative $\mathrm{D}_vf(x)$ exists for all $v\in \mathrm{T}_x(\mathbb{R}^d)$ and (ii) the map $\mathrm{T}_x(\mathbb{R}^d)\ni v \mapsto \mathrm{D}_vf(x)\in \mathbb{R}$ is linear and continuous. (You actually get continuity for free, but I include it because it belongs there if you replace $\mathbb{R}^d$ with something infinite dimensional.) In this case, I write $v^a\nabla _af(x):=\mathrm{D}_vf(x)$. A covariant tensor field $T_{a_1\ldots a_l}$ is differentiable at $x\in \mathbb{R}^d$ iff $[v_1]^{a_1}\cdots [v_l]^{a_l}T_{a_1\ldots a_l}$ is differentiable for all $[v_1]^a,\ldots ,[v_m]^a\in \mathbb{R}^d$, in which case the derivative is the tensor field of covariant rank $l+1$, $\nabla _aT_{a_1\ldots a_l}$ that sends $l+1$ vectors $v^a,[v_1]^a,\ldots ,[v_l]^a$ to $v^a\nabla _a\left( [v_1]^{a_1}\cdots [v_l]^{a_l}T_{a_1\ldots a_l}\right)$. (Of course, you use essentially the same definition for arbitrary tensors, but math.stackexchange is not letting me stagger indices and I don't need to know how to differentiate contravariant tensors for the purposes of this question).

The question is:

If $f$ is twice differentiable at $x$, must we have that $\nabla _a\nabla _bf(x)=\nabla _b\nabla _af(x)$?

The definition given above is strictly weaker than Fréchet differentiability, but strictly stronger than Gâteaux differentiability (if you want the examples that show strictness, ask in the comments). The result is true for the Fréchet derivative (see, e.g., Pugh pg. 281), but false for the Gâteaux derivative (see, e.g., the standard example on Wikipedia). This is not a counter-example for the definition above because the second derivative fails to be linear at the origin (and so it is not considered to be twice differentiable there). Furthermore, it is true for all three definitions if we require continuity of $\nabla _a\nabla _bf(x)$ at $x$. But can we remove this continuity hypothesis?

For what it's worth, I think it is false.