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I know that for $1 \leq p < \infty$ we have $C_c(X)$ dense in $L^p(X,\mu)$. In $L^p(X,\mu)$ the evaluation functional is not bounded, but in $C_c(X)$ it is.

So if I define $T_x : C_c(X) \to \mathbb{R}$ as $T_x f = f(x)$ for $x \in X$ I can extend this functional to $L^p(X,\mu)$ using the Hahn Banach (HB) theorem.

To make sure I understand how to possibly apply HB I wonder:

  1. Is my application correct?
  2. Am I right when I say that the extension if $T_x$ is bounded?

Thank you.

user8469759
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3 Answers3

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You cannot extend $T_x$ to a bounded linear functional via Hahn-Banach because, in general, the evaluation functional in $(C_c(X), ||\cdot||_p)$ is not bounded. For example, let $X = [0,1]$ and let $$f_n(x) = \begin{cases} (2n - 2n^2x)^{1/p} & 0 \leq x \leq \frac{1}{n} \\ 0 & \text{o.w.} \end{cases}$$ Observe that $||f_n||_p = 1$ for all $n$. Now let $\delta_0$ be the point evaluation functional at $0$ and observe that: $$\delta_0(f_n) = (2n)^{1/p} \to \infty$$ As $n \to \infty$.

Jose Avilez
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You can linearly extend $T_x$ to $L^p$ and you don't even need Hahn-Banach for that. But to extend a linear functional from a subspace $F\subset E$ to a bounded linear functional on $E$, you need it to be continuous on $F$ to begin with - continuous with respect to the norm of $E$, that is. But the functional $T_x$ is in general not continuous with respect to the $L^p$ norm on $C_c(X)$, so there is no chance to extend $T_x$ to a bounded linear functional on $L^p(X,\mu)$.

Of course, things also depend on the measure space. If each point has positive measure, then point evaluation is bounded as pointed out by David C. Ullrich in the comments. Moreover, the measure $\mu$ needs to have full support and give finite mass to compact sets in order to view $C_c(X)$ as a subspace of $L^p(X,\mu)$.

MaoWao
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  • How do I construct the extension? I am missing that bit. – user8469759 Jul 12 '21 at 14:02
  • That's essentially (infinite-dimensional) linear algebra: Take a (vector space) basis of $C_c(X)$, extend it to a basis of $L^p(X,\mu)$ and define the value of the extended functional on the basis elements in $L^p\setminus C_c$ by whatever you like. – MaoWao Jul 12 '21 at 14:11
  • I don't know a basis for $C_c(X)$, I know basis mainly for $L^2$ spaces. How do I extend it to a basis in $L^p(X,\mu)$? – user8469759 Jul 12 '21 at 14:46
  • Well, no one "knows" a basis of $C_c(X)$ unless in some special situations (say when $X$ is discrete). But Zorn's lemma guarantees that it exists and that it can be extended to a basis of $L^p(X,\mu)$. – MaoWao Jul 12 '21 at 15:29
  • Can I have a reference to that statement? Cause I know the existance of a basis theorem in linear algebra, finite dimensional. – user8469759 Jul 12 '21 at 15:54
  • See these notes for example: http://www.math.lsa.umich.edu/~kesmith/infinite.pdf Note that what is actually proven there is that one can extend each linearly independent set to a basis, which you can use to extend a basis of $C_c(X)$ to a basis of $L^p(X,\mu)$. – MaoWao Jul 12 '21 at 16:02
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Yes, you can use Hahn-Banach to extend this functional to all of $L^p(X,\mu)$. However, since $T_x$ is not bounded, the extended functional is linear, but not continuous. Moreover, the extension is not unique. And for $u \not\in C_c(X)$, $T_x u$ might not be related to $u(x)$.

Putting everything together: Yes, you can do it, but it is not a good idea. Don't try this at home and wear safety goggles ;)

gerw
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  • I thought in order to extend using HB you need the functional to be bounded, don't I? – user8469759 Jul 12 '21 at 13:49
  • Yes and no: HB needs a sublinear functional defined on the entire space which bounds the linear functional. However, the proof of HB works also if you neglect the sublinear functional and you can just extend the linear functional to the entire space. – gerw Jul 12 '21 at 14:51