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The norm of an element $(\gamma_k)=x \in l^{\infty}$ is defined to be $\sup_k \gamma_k <\infty$. The norm of an element $(\gamma_k)=x \in l^p$ is defined to be $(\sum_k \gamma_k^p)^{1/p} < \infty$, even for $p=1$!

I can't see how $\displaystyle{\lim_{p \to \infty}} (\sum_k \gamma_k^p)^{1/p} = \sup_k \gamma_k$ holds?

And, if they just are separate definitions, then why the case '$p=\infty$' is in the 'category' of $L^p$ spaces?

Added - Understanding answers to this question requires a knowledge more than an undergraduate Analysis, i.e. my current level. Simple detailed answers would be much appreciated.

levap
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    Why do you mean by limit of a (measure? metric? vector?) space? – Fimpellizzeri Aug 27 '17 at 23:31
  • @Fimpellizieri, the second paragraph of OP. –  Aug 27 '17 at 23:31
  • You mean the norms then -- should probably edit the question title to reflect that. – Fimpellizzeri Aug 27 '17 at 23:32
  • @JohnMa, yes this is duplicate but the answers there are difficult for an undergraduate level math.. though I didn't see that question before I posted this. –  Aug 27 '17 at 23:37
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    It seems confusing to me in that on one hand $L^p$ is generally used for spaces of functions, but the way you write the norms in terms of series points toward $\ell^p$, the spaces of sequences. – Fimpellizzeri Aug 27 '17 at 23:56
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    @Edi, Yes, I thought about that too. But on the other hand I cannot think of another method besides though given there. You just need to change $X = \mathbb Z$ and $\int$ to $\sum$. –  Aug 28 '17 at 00:00
  • $\ell^p=L^p((\mathbb{N},\mathcal{P}(\mathbb{N}),c))$ where $c$ is the counting measure :) – Ian Aug 28 '17 at 01:17

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In general if $f \in L^p$ for some $1 \leq p<\infty$ and $f \in L^\infty$ then $f \in L^q$ for all $q \in [p,\infty]$. In this case $\| f \|_{L^\infty} = \lim_{p \to \infty} \| f \|_{L^p}$. This is easiest to see in the case of $\mathbb{R}^n$, where (assuming $x_i \neq 0$)

$$\| x \|_p = |x_i| \left ( \sum_{j=1}^n \left ( \frac{|x_j|}{|x_i|} \right )^p \right )^{1/p}$$

If $|x_i|$ is the biggest, then all the terms either go to zero or $1$. In either case since $n$ is fixed,

$$\left ( \sum_{j=1}^n \left ( \frac{|x_j|}{|x_i|} \right )^p \right )^{1/p} \to 1$$

as $p \to \infty$. The idea in the general case is that the condition $f \in L^p$ ensures that $\left ( \frac{|f(x)|}{\| f \|_{L^\infty}} \right )^p$ goes to zero rapidly with $p$ for "most" $x$.

Ian
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