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I'm trying to do this problem but I don't know:

Show that, considering the continuous functions on $\mathcal{C}([0, 1])$ as a subset of $L_1([0, 1])$, the linear functional on this subset $f\mapsto f(\frac{1}{2})$ is not bounded.

I have tried to find a sequence of functions $\{f_n\}$ such that, if $K$ is that functional, then $\lim \dfrac{\|Kf_n\|_1}{\|f_n\|_1}\to \infty$.

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What have you tried? –  wj32 Oct 30 '12 at 11:54
    
I have tried to find a sequence of functions $\{f_n\}$ such that, if $K$ is that functional, then $\lim \dfrac{\|Kf_n\|_1}{\|f_n\|_1}\to \infty$. –  Orkidea Oct 30 '12 at 11:57

3 Answers 3

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Hint. Consider positive functions with peak of height $1$ at the point $1/2$ with small integral. For example $$ f_n(t)=\max\left(0, 1-\left|n\left(t-\frac{1}{2}\right)\right|\right) $$

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I think you meant that the $+$ in your formula be a $-$. –  brom Oct 30 '12 at 12:26
    
I think you have a $t$ that should be an $x$ - or vice versa. –  kahen Oct 30 '12 at 12:27
    
Thanks for your suggestions –  userNaN Oct 30 '12 at 12:31
    
If I have done my calculations, it gives $\|f_n\|_1=\dfrac{1}{n}$ and $\|Kf_n\|_1=1$, so it works. Thanks. –  Orkidea Oct 30 '12 at 13:44

Continuing a bit your try: take each $f_n$ nonnegative and such that $f_n(\frac 12)=1$. Then to get your limit to be infinite you just have to make the area under the graph of $f_n$ going to zero.

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Take $f_n$ to be supported on an interval of length $2^{-n}$ containing $\tfrac12$ and with $\lVert f_n\rVert_\infty = f(\tfrac12) = 2^n$. For concreteness, you could take a "spike" such that the area under the graph is just a triangle.

The $f_n$ form a $\lVert\cdot\rVert_1$-bounded set, but their image isn't bounded in $\mathbb R$.

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This isn't really worth editing the answer for, so I'll just stick it in a comment: A linear map between normed spaces is continuous if and only if it takes bounded sets to bounded sets. I believe this is (part of) the reason why continuous linear maps are called bounded maps. –  kahen Oct 30 '12 at 12:48

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