I would like to solve the following problem
If $J$ is a functional twice differentiable from a normed space to $\mathbb{R}$, prove that
$$ J(u+w) = J(u) + J’(u)w + \frac{1}{2}J’’(u)(w,w) + o(\lVert w\rVert^{2}) $$
Where $\lim_{w\to 0}\frac{o(\lVert w \rVert^{2})}{\lVert w\rVert^{2}} = 0$
First I notice that they identify the second derivative (which is in $\mathcal{L}(V,\mathcal{L}(V,R))$) with a continuous bilinear form.
From this, by the fact that $J$ is differentiable and $J’’(u)(w,w)$ is a continuous bilinear form we have
$$ 0\leq\lvert J(u+w) - J(u) - J’(u)w\rvert + \frac{1}{2}C\lVert w\rVert^{2} $$
For some constant $C$. However if I divide by $\lVert w\rVert^{2}$ I cannot use the differentiability and for the second member I will get $\frac{1}{2}C$ which is not of interest since I want to show that it goes to $0$.
I think this is really not difficult but I am stuck. I would like to have your help through some hints if possible please.
Thank you a lot !
Here is an update thanks to Didier, hope it is correct.
Let $w\in E$ different from the null vector otherwise there is nothing to do. Take $v=\frac{w}{\lVert w\rVert}$ and consider the map $g(t)=J(u+tv)$
We can notice that $g’(t) = J’(u+tv)(v)$ and $g’’(t) = J’’(u+tv)(v,v)$.
Now by the usual mean value theorem we have the following expression (there is no problem at $0$ since $J(u)$ is differentiable)
$$ g(t) = g(0) + g’(0)t + \frac{t^2g’’(0)}{2} + o(t^2) $$
Which means that $\forall\epsilon>0$ , $\exists\delta>0$ such that $ 0<t^2\leq\delta$ implies
$$ \frac{\lvert g(t) - g’(0)t - \frac{t^2g’’(0)}{2}\rvert}{t^2}\leq\epsilon $$
We see that $J(u+ w) = g(\lVert w\rVert)$. Thus take $t=\lVert w\rVert$, take $\epsilon>0$, then there exists $\delta>0$ such that $\lVert w\rVert^{2}\leq \delta$ implies $$ \frac{\lvert J(u+w) - J’(u)(w) - \frac{J’’(u)(w,w)}{2}\rvert}{\lVert w\rVert^2}\leq\epsilon $$
Consider the map \begin{array}{r|ccc} g \colon & \Bbb R & \longrightarrow & \Bbb R \\ & t & \longmapsto & J(u+tv), \end{array} where $\|v\|=1$. Then $g$ is a real function that is twice differentiable at $0$, and there exists a function $h\colon \Bbb R \to \Bbb R$, with $h(t) \longrightarrow_{t\to 0} 0$, such that $$ \forall t \in \Bbb R, \quad\left|g(t) - g(0) - tg'(0) - \frac{1}2t^2g''(0) \right| \leqslant h(t) t^2. $$ Now, notice that $g(0) = J(u)$, $tg'(0) = J'(u)tv$ and $t^2g''(0) = J''(u)(tv,tv)$. Set $w = tv$, so that $t= \|w\|$. Can you go on from there?
(NB: one needs to be careful and to show that $h$ can be chosen independently of $v$.)