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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.

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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 \[ \|x^*\| = \sup_{x \in B_X} x^*(x) = -\inf_{x\in B_X} x^*(-x) = -\inf_{x\in B_X} x^*(x). \]

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Sorry, I think if you put, for example, f(t)=t if t<=1/2 and f(t)=1-t if t>=1/2, then $\Gamma(t)$ it isn't equal to $B_X$. –  Manolete Jul 4 '12 at 20:23
    
@Manolete I think there should be $f\in\Gamma$ instead of $t\in\Gamma$ –  no identity Jul 4 '12 at 20:26
    
@Norbert Thx. You're right. –  martini Jul 4 '12 at 20:56
    
Ah, ok. I have read it bad. So it's clear that $\Gamma(t)=B_X$. Then $\inf\limits_{f\in \Gamma} x^*\big(f(t)\big)=\inf\limits_{f(t)\in \Gamma(t)} x^*(f(t))=\inf\limits_{x\in B_X} x^*(x)$. Thanks a lot! –  Manolete Jul 4 '12 at 21:41

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