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I have to prove that the sequence $f_n(x)=x^n$ does not weakly converge in $(C^0([0,1]), \|\cdot\|_\infty)$.

Some hint?

2 Answers 2

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It fails to converge pointwise to a continuous function. Point-evaluations are bounded linear functionals on your space.

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    Sorry, but I don't understand. For every $x \in (0,1)$ we have $x^n \to 0$, isn't it pointwise convergence? – 2017-01-12
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    @Bobech, you wanted a hint not a full solution. If it converged weakly, it would have converged pointwise (to a continuous function) too, but it actually does not. The pointwise limit of your sequence is discontinuous. – 2017-01-12
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    ok, now it's clear. Thanks – 2017-01-12
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Hint: consider the functional $$f \mapsto f(t)$$ for all $t \in [0,1]$.

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    Let's call this functional $\phi_t$. So we get $|\langle \phi_t, f_n\rangle |=|\phi_t (f_n)|=|t^n|$ and it does not converge for $n \to \infty $ only for $t=1$. Is it suffice? – 2017-01-12
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    No, it converges for all $t$, but the limit does not depend continuously on $t$. – 2017-01-12
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    Sorry, it was 'not converge to 0' – 2017-01-12