let $E$ be a banach space and $f:E\rightarrow \mathbb{R}$ differentiable
we know that $x_0$ is local minimizer , if there existe a neighborhood $ V\subset E$ such that :
$f(x_0)\leq f(x)$ for all $x\in V$
and a way to prove $x_0$ is local minimizer is to prove that $f^{'}(x_0)= 0$ and $f^{"}(x_0)>0$
I have a little confusion : I believe that if I can show that $f^{'}(x_0)= 0$ and $f^{"}(x_0)\geq0$ we can easily see by Taylor expansion that $f(x_0)\leq f(x)$
I do not see where he is the problem
thank you @humanStampedist your contre example is great ,
to simplify the notations i will take $ x_0=0$ and $F=f$ so
$F(t)=F(0)+ t F^{'}(0) +\frac{t^2}{2}F^{"}(0) + o(t^2)$ and we have $F'(0)=0$ so $F(t)=F(0)+\frac{t^2}{2}F^{"}(0) + o(t^2)$
as $F^{"}(0)\geq 0 $ then $\frac{t^2}{2}F^{"}(0) + o(t^2)$
this imlpies $F(t)=F(0)+\frac{t^2}{2}F^{"}(0) + o(t^2)\geq F(0)$
in this cases 0 is a local minimum