I was reading Shifrin's "Multivariable calculus", and stuck upon the proof of the symmetry of Hessian matrix. I am really not happy with it as it requires introduction of coordinates, so it seemed natural to me to try to introduce the Hessian in coordinate-free way as follows:
We say, that a function $f\colon V\to W$ between Euclidean spaces is twice differentiable at $v\in V$ if it is differentiable in a neighbourhood of $v$, and there exists a bilinear map $D_2f(v)\colon V\times V \to W$ and for every $\epsilon > 0$ there exists $\delta>0$ such that for every $h,k\in V$ with $\|\:h\|\:,\|\:k\|\: < \delta$ we have $$\frac{\|\:Df(v+h)k - Df(a)k - D_2f(v)(h,k)\|\:}{\|\:h\|\:\|\:k\|\:} < \epsilon.$$
So the question is to prove the symmetry of $D_2f(v)$. A natural way would be to estimate $$A = \frac{ \|\:D_2f(v)(h,k) - D_2f(v)(k,h)\|\:}{\|\:h\|\:\|\:k\|\:}.$$
So we fix $\epsilon > 0$. By hypothesis, we can find $\delta$ such that for any $h,k$ of norm less than $\delta$ we have $$ A < 2 \epsilon + \frac{ \|\:Df(v + h)k - Df(v)k - Df(v + k)h - Df(v)h\|\:}{\|\:h\|\:\|\:k\|\:} $$
And now we try to use the definition of the derivative:
For any $\epsilon >0$ there exists $\delta > 0$ such that for any $h\in V$ with $ \|\:h\|\:< \delta$ we have $$\frac{\|\:f(v + h) - f(v) - Df(v)h\|\:}{\|\:h\|\:} < \epsilon.$$
Decreasing $\delta$ if needed we use triangle inequality four times and conclude $$ A < 2\epsilon + 2\frac {\epsilon}{\|\:h\|\:} + 2\frac {\epsilon}{\|\:k\|\:}$$ (all the remaining terms vanish identically).
So the question is if it is possible to get rid of the norms of $h$ and $k$ in the denominators?