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Let $ X = C[0,1] $ with the supremum norm $ || \cdot ||_\infty $. Consider the functional $ \phi $ defined for $ f \in X $ by $$ \phi(f) = \int_0^\frac{1}{2} f(x) dx - \int_\frac{1}{2}^1 f(x) dx .$$ Show that if $ f \in X $ with $ ||f|| = 1 $ then $ |\phi(f)| < 1 $.

A previous part of this question was to show that $ \phi $ is a bounded linear functional with norm 1 which I have done already. But I do not know how to show that $ |\phi(f)| < 1 $. Any help or hints would be appreciated.

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Prove by contradiction. Suppose $|\phi (f)|=1$. Then $1=|\int_0^{1/2}f-\int_{1/2}^{1}f| \leq \frac 1 2 +\frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f \equiv 1$ or $f \equiv -1$. But then $\phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].

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