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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 $\lim \dfrac{\|Kf_n\|_1}{\|f_n\|_1}\to \infty$.

Orkidea
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  • 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 Answers3

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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.

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

Norbert
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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 $\mathbb R$.

kahen
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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