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Given a $g$ an odd function the question is to exhibit a continuous linear functional from $C[-1,1]$ $\phi$ such that

$|\phi|=1$ , $\phi(g)=|g|_\infty$ and $\phi$ is zero on the even functions.

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Can you not just first define the functional on the subspace generated by all even functions and your fixed odd function $g$ -- it should be clear that there is a unique way to do this -- check that it is bounded, and then use an easy version of Hahn-Banach to extend the domain to $C[-1,1]$ without increasing the norm? – Pete L. Clark Feb 29 '12 at 23:20
up vote 4 down vote accepted

Let $x_0$ be a point where $g(x_0)=\Vert g\Vert_\infty$ (as $g$ is an odd function, such a point exists). Define $\phi(f)={1\over2}\bigl(f(x_0)-f(-x_0)\bigr)$.

Then $\phi$ linear. The norm of $\phi$ is 1 by virtue of the fact that $\phi(g)=\Vert g\Vert_\infty$ and $|\phi(f)|\le {1\over 2}(|f(x_0)|+|f(-x_0)|)\le\Vert f\Vert_\infty$. Also, if $f$ is even, $\phi(f)=0$.

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