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I have the following question:

Consider the normed space $(C([0,1]),||\cdot||_{\infty})$. Let $x_{0}\in [0,1]$ and define $F:C([0,1])\to \mathbb{R}$ by


Show that $F$ is discontinuous with respect to $||\cdot||_{1}$.

I realise that I simply need to find a counterexample where $f_{n}\to f$ with respect to the one-norm, but where $F(f_{n})\not \to F(f)$. However, I see no way in which to do this. I hate being lazy with this so would appreciate a small push in the right direction.

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There are some very simple, familiar functions $f_n$ that are $0$ at $0$, $1$ at $1$, and squeezing towards the $x$-axis as $n$ increases. They give you a nice example when $x_0=1$. For other values of $x_0$ you can adapt the basic idea of that family, most easily by using piecewise linear functions. –  Brian M. Scott Mar 8 '12 at 1:45

2 Answers 2

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If I understand your question and you mean $L^1$ norm (you also mention sup norm and $F$ is continuous with respect to that, which is perhaps point of problem), you could let $f_n$ be $0$ on $[0,1/2-1/n^2]$, the line from $0$ to $n$ on $[1/2-1/n^2,1/2]$, the line from $n$ to $0$ on $[1/2,1/2+1/n^2]$, and $0$ on $[1/2+1/n^2]$. Then $f_n \rightarrow 0$ in $L^1$, but $F(f_n)=n$.

Basically any function that is zero except for a "high spike of small area" will work.

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Here is an idea. Squeeze a bump.

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-1, you could keep these ideas that are relatively small enough for an answer as a comment. –  user21436 Mar 9 '12 at 1:08
Actually that is an answer. I am trying to avoid "giving away the shop." –  ncmathsadist Mar 9 '12 at 1:34

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