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Is there a norm in the C[0,1] space function such that this happens?

$ f_n(x) \rightarrow f(x) $ if and only if $ \|f_n - f\| \rightarrow 0$

Whene $f_n$ is a function sucession, and $ f_n(x) \rightarrow f(x) $ means Pointwise convergence

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A norm on what space of functions? And is $f_n(n)$ a typo? – Nate Eldredge Sep 19 '12 at 1:16
Assuming you mean $f_n(x)$ rather than $f(x)$, this is true by definition, in the sense that $f_n\to f$ in the norm topology induced by $\|\cdot\|$. To make the question meaningful you need to specify the topology in which you want $f_n\to f$. – Alex Becker Sep 19 '12 at 1:18
"Puntal convergence" is very amusing, and practically a googlewhack to boot. It sure sounds like you mean "pointwise", so I edited it to that. – rschwieb Sep 19 '12 at 1:22
@rschwieb so now what is a googlewhack? Meanwhile a punt is a type of boat or a method of kicking a ball. Pointwise also works. – Will Jagy Sep 19 '12 at 1:24
Take a look at this: – Jack Sep 19 '12 at 1:33
up vote 7 down vote accepted


Suppose there were such a norm; call it $\|\cdot\|$. For each $n$, let $f_n$ be any nonzero continuous function supported in $(0,1/n)$. Since $f_n$ is not the zero function, we must have $\|f_n\| > 0$. So if we let $g_n = \frac{1}{\|f_n\|} f_n$, then we have $\|g_n\| = 1$ for all $n$; in particular, $g_n$ does not converge to zero in the norm $\|\cdot\|$. But $g_n$ is a continuous function supported in $(0,1/n)$, so $g_n(x) \to 0$ pointwise. This is a contradiction.

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thanks I was looking for that it is perfect. – Maria Sep 19 '12 at 1:47

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