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There is a norm $\Vert .\Vert$ in the space $C([a,b],\mathbb{R})$ such that

convergence pointwise implies convergence in norm $\Vert .\Vert$ ?

I think not because if there would be the natural candidate standard $C([a,b],\mathbb{R})$

any suggestion is welcome.

helmonio
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  • The answer is no. This is a duplicate see http://math.stackexchange.com/questions/33476/norm-for-pointwise-convergence – Omran Kouba Apr 17 '14 at 21:13
  • @OmranKouba This question asks for a norm weaker than pointwise convergence, the other one for equivalence. – Daniel Fischer Apr 17 '14 at 21:16
  • @DanielFischer, Yes, you are right, I think, I already found a distance, based on a previous post: $d(f,g)=\int_a^b\frac{\vert f-g\vert }{1+\vert f-g\vert}$. – Omran Kouba Apr 17 '14 at 21:31
  • Distance, but not a norm, @OmranKouba. I suspect there is no norm weaker than pointwise convergence, but I don't see how to prove it. – Daniel Fischer Apr 17 '14 at 21:33
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    By "convergence pointwise implies convergence in norm", do you mean general convergence (of filters/nets), or only convergence of sequences? – Daniel Fischer Apr 18 '14 at 01:11

1 Answers1

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There is no such norm. In fact, for every norm on $C^\infty([0, 1])$ there is a sequence of smooth functions that converges pointwise but not in the norm.

Let $\Psi$ be a bump function on $\mathbb{R}$ with support $[0,1]$ and define $\{f_n\}_{n=1}^\infty$ by $f_n(x) = \Psi(nx)$. Although $f_n$ is never identically zero, it is easy to verify that for every $x$ the sequence $\{ f_n(x) \}$ is eventually zero, so $f_n \to 0$ pointwise.

If we fix a norm, we can define $g_n = nf_n/\|f_n\|$. Since $g_n$ is zero wherever $f_n$ is, we also have $g_n \to 0$ pointwise. On the other hand, $\|g_n\| = n$ is unbounded, so $\{g_n\}$ is not even a Cauchy sequence in the norm.

Niels J. Diepeveen
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