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6 votes

The space $C_c$ of real-valued compactly supported, continuous functions is not a Banach space under any norm

If you assume the axiom of choice, this is false. Indeed, notice that $\mathcal C_c(\mathbb R)\supset \mathcal C_0([0,1]),$ hence $\text{dim}(\mathcal C_c(\mathbb R))\ge \mathfrak c$. On the other ...
Pelota's user avatar
  • 1,003
1 vote

Is this a valid criterion for Bochner integrability?

Yes, at least if $f$ is strongly measurable. For each $n$, let $g_n$ be a simple function such that $\int\|g_n-f_n\|\mathrm d x< 1/n$. Let $\epsilon>0$. For $n$ large enough, $\int\|f_n-f\|\...
Michael Greinecker's user avatar
1 vote

(Copy) Set of linear functionals span the dual space iff intersection of their kernels is {0} .

Consider the space $\Bbb F^{(\Bbb N)}$ of lists $(a_1,a_2,a_3,\dots)$ of elements of $\Bbb F$ such that $\{n \in \Bbb N : a_n \neq 0\}$ is finite. This space has dimension $|\Bbb N|$, since a basis ...
azif00's user avatar
  • 21k
1 vote

Subdifferential of supremum norm on a space of continuous functions

I think it is true. As commented by Evangelopoulos Foivos, by the Banach–Alaoglu and Krein–Milman theorems it suffices to show that $S$ coincides with the set $\operatorname{ext}(\partial \| f \|_\...
user486506's user avatar
1 vote

For $1\le p < +\infty$ $L^p$ is a Banach space: Real and abstract analysis, Hewitt - Stromberg

For your first question, what you did to find the first two terms was fine. Since you started the argument, you may as well finish it and describe how to find all of the terms. You can find a strictly ...
Dean Miller's user avatar
  • 1,934

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