I really appreciate your help in checking the below proof (especially the last section).
Let $(E_{i})_{i=1 \dots n}$ and $F$ be normed vector spaces and $f: \prod{E_{i}}\rightarrow F$ a twice-differentiable function (no hypothesis is made on the continuity of partial derivatives). We want to prove that $\forall a,h,k \in \prod{E_{i}}, f''(a)(h)(k)=f''(a)(k)(h)$.
We write $\Delta(a,h,k)=f(a+h+k)-f(a+h)-f(a+k)+f(a)$. Then:
$||\Delta(a,h,k) – f''(a)(k)(h)|| \leq$
$||\Delta(a,h,k) – f'(a+k)(h)+f'(a)(h)||+||f'(a+k)(h)-f'(a)(h)-f''(a)(k)(h)||$
Let $(1)$ denote the first summand and $(2)$ the second.
I can show that $(2)=o(||h||.||k||)$ using the definition of $f''$ therefore I will now try to do the same with $(1)$ (from there we can conclude by applying the same reasoning to $||\Delta(a,h,k) – f''(a)(h)(k)||$ and using the linearity of the derivatives: I am not interested in that part of the proof).
$(1) = ||\Delta(a,h,k) – f'(a+k)(h)+f'(a)(h)||=$
$||\delta(a,h,k)-\delta(a,h,0)|| \leq $
$||k||.Sup_{x \in B(0,||k||)}{||f'(a+h+x)-f'(a+x)-f''(a+x)(h)||}$
where $\delta(a,h,k)= f(a+h+k)-f(a+k)-f'(a+k)(h)$, $B(0,||k||)$ the closed ball with center $0$ and radius $||k||$, and using the Mean Value theorem.
Since $f'$ is differentiable it is continuous and the $Sup$ is reached at some point say $x0$.
$(1) \leq ||k||.||f'(a+h+x0)-f'(a+x0)-f''(a+x0)(h)|| \leq $
$||k||.||f''(a+x0)(h)|| \leq$
$||k||.||h||.||f''(a+x0)|| = o(||h||.||k||)$
P.S.: Is there any useful, simple way to weaken the hypotheses?
Best Answer
I would start with $$ (1) \|\Delta(a,h,k) - f'(a+k)(h)+f'(a)(h)\| = \| d(a,h,k)-d(a,0,k)\| $$ with $$ d(a,h,k) =f(a+h+k) - f(a+h) - f''(a)(k,h). $$ To apply the mean value theorem we need the derivative of $a$ with respect to $h$ $$ \partial_h d(a,h,k) = f'(a+h+k) - f'(a+h) - f''(a)k. $$ Now it is tempting to apply mean value theorem again. We resist and write this expression as $$ \partial_h d(a,h,k) = f'(a+h+k) - f'(a+h) - f''(a)k\\ = ( f'(a+h+k) - f'(a) - f''(a)(h+k) ) - ( f'(a+h)-f'(a)-f''(a)h)\\ =: e(a, h+k) -e(a,h) $$ with $$ e(a,h):=f'(a+h)-f'(a)-f''(a)h. $$ Now both terms $e(a, h+k)$ and $e(a,h)$ are of order $o(\|h\|+\|k\|)$, and $\partial d_h$ is of the same order as well. This shows that (1) if of order $o(\|h\|^2+\|k\|^2)$.
The main point of the proof is the avoid any mean value theorem for $f'$ and avoid values of $f''$ at points different from $a$. This proof only uses twice Frechet differentiability of $f$ at one point $a$. No continuity of derivatives is needed.