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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?2012-09-19
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    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 $f_n\to f$.2012-09-19
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    "Puntal convergence" is very amusing, and practically a googlewhack to boot. It sure sounds like you mean "pointwise", so I edited it to that.2012-09-19
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    @rschwieb http://en.wikipedia.org/wiki/Puntal so now what is a googlewhack? Meanwhile a punt is a type of boat or a method of kicking a ball. Pointwise also works.2012-09-19
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    What properties does your $(f_n)$ have?2012-09-19
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    Sorry yes is Pointwise ahah i miss translate, no the $f_n$ doesnt hold any particular properties.2012-09-19
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    Take a look at this: http://math.stackexchange.com/q/33476/94642012-09-19
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    @WillJagy Can't tell if your question was rhetorical, so: to the best of my knowledge, a googlewhack is managing to get only a single hit with a one or two word search phrase in google. When I tried "puntal convergence" I only got two hits, both here.2012-09-19
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    @rschwieb, that makes sense. By itself "puntal" turned up a few things, but when I looked up googlewhack it did say something about two-word combinations.2012-09-19

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No.

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.2012-09-19