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Norm for pointwise convergence

Let $V=C([0,1],\mathbf{R})$ be the vector space of continuous real-valued functions on $[0,1]$.

Let $(f_n)$ be a sequence in $V$. Then $(f_n)$ converges with respect to the max-norm on $V$ if and only if $(f_n)$ is uniformly convergent.

As a consequence, since the limit of a uniformly convergent sequence in continuous, we conclude that $V$ endowed with the max-norm is a Banach space.

Now, on $V$ there is also a notion of pointwise convergence. Is there a norm $\Vert \cdot \Vert_{pc}$ on $V$ such that a sequence $(f_n)$ in $V$ converges with respect to $\Vert \cdot\Vert_{pc}$ if and only if $(f_n)$ is pointwise convergent?

Note that $V$ will not be Banach under this norm.

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marked as duplicate by Byron Schmuland, t.b., Rasmus, Jonas Teuwen, Asaf Karagila Oct 20 '11 at 0:22

This question was marked as an exact duplicate of an existing question.

That anwers my question. Should have looked around a bit more carefully. So there is no norm on $V$ in which convergence is pointwise convergence. – Bana Oct 19 '11 at 19:04
No problem, and welcome to Math Stackexchange. – Byron Schmuland Oct 19 '11 at 19:06

No, $V$ it self is too small. Look at $$f_n(x)=\max(1,1/(n\cdot x)),\qquad \qquad n=1,2,\ldots$$ it converges pointwise, but the limit is not in $V$.

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This does not exclude the possibility of norming the topology of pointwise convergence. It only shows that there can't be a complete norm. In this thread I show that the topology is neither normable nor metrizable. – t.b. Oct 19 '11 at 18:54
Why is an uncountable product not even metrizable? Is this an easy fact? – Bana Oct 19 '11 at 19:05
@Bana: Show that it is not first countable. – t.b. Oct 19 '11 at 19:15

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