Proving $\inf\limits_{f\in\Gamma} \{ F(y(t))(f(t))\}= -\|F(y(t))\| $ I would like to proof the next claim:
Let $X$ a Banach space, $F\colon X\to X^*$ a linear continuous function, 
$$
\Gamma:=\{f\in (\mathcal{C}([0,1],X)\,:\, f(0)=f(1)=0\mbox{ and }\|f\|\leq 1\}
$$ and fix $t\in [0,1]$, $y\in \Gamma$. Then $F(y(t))\in X^*$. How I can prove that 
$$
\inf\limits_{f\in\Gamma} \{ F(y(t))(f(t))\}= -\|F(y(t))\| 
$$
? I tried it using Hahn-Banach, but maybe there is another easier way. Thanks in advance.
Edit: sorry, I forgot to put that F es linear... Thanks for the answers.

You need to assume $F(0) = 0$ for this to hold as the left hand side is $0$ for $t\in \{0,1\}$, hence the right hand side should also be zero, i. e. $F(0) = 0$. As $F$ and $y$ are evaluated at exactly one point, let's write $x^* = F\bigl(y(t)\bigr)$, if $t \in (0,1)$, then $\Gamma(t) = \{f(t)\mid f\in \Gamma\} = B_X$, so you have to prove $\inf_{x\in B_X} x^*(x) = -\|x^*\|$, which holds as 
\[
\|x^*\| = \sup_{x \in B_X} x^*(x) = -\inf_{x\in B_X} x^*(-x) = -\inf_{x\in B_X} x^*(x).
\]
