Any function $f$ in $L^2$ is a actually an equivalence class and has properties that only hold "almost everywhere." But it would be convenient to speak of the value of $f$ at certain points like $f(0)$.
Is there a meaningful way of defining this?
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The more natural way would be to define $f(0)$ as an average of $f$ near 0:
(Re)define $f(0)$ to be the limit as $\varepsilon$ tends to 0 of $$\frac{1}{\mu(B(0,\varepsilon))}\int\chi_{B(0,\varepsilon)}(x)f(x)d\mu(x).$$ provided this limit exists (and it does a.e. (at least) by the Lebesgue Differentiation Theorem. In this context when this limit exists it is said that 0 is a Lebesgue point, and what the Lebesgue Differentiation theorem says is that almost every point is a Lebesgue point). This way you're giving $f$ the value you'd give it if you could paint its plot. Think for example in the function $\chi_{\mathbb{Q}\cap[-1,1]}$. However, this won't work if 0 is not a Lebesgue point (take for example the Heavyside function in $[-1,1]$, $\chi_{[0,1]}$)
So this approach won't work in general. Nevertheless notice that this would not be useful in the theory of $L^p$ functions cause even if you could do this with every point you would end up with a function that would differ from the original one in a set of measure 0, so as a $L^p$ function it would be exactly the same.
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2I think you likely want to normalize this by $1/\mu([-\epsilon, \epsilon])$, but there's still a huge problem: For an $L^2$ function this may not exist, since differentiation theorems give only almost everywhere convergence. – Jun 20 '16 at 01:25
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2@copper.hat Yes right!? Lol, always found hilarious that fast trigger of some people when it comes to downvote. But, who cares about points? Always pleased to find people who understand this is about learning math... – Jun 20 '16 at 01:36
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You might want to mention that the points where this definition works are called the Lebesgue points. But the problem with the definition is essentially that it is an almost everywhere true assertion, so you could be in trouble for a set of zero measure. I know you have actually said all this glibly but precisely (aside from the name "Lebesgue points") but it could be rephrased for those who don't read carefully enough (even though they're happy to downvote). Also perhaps replace "differs from the original one" with "could differ ...."; this then makes for a good question with a good answer. – Selene Routley Jun 20 '16 at 01:51
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@T.Bongers I think Jonh has taken this into account in "Provided the limit exists ... it does a.e."and also in his/her last sentence (although i'd write "could differ" rather than "differ"). It could perhaps be rephrased for clarity, but it is stated nonetheless. – Selene Routley Jun 20 '16 at 01:56
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@Wet My point is that getting an object that exists a.e. doesn't resolve the issue - it's circular. – Jun 20 '16 at 02:00
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@T.Bongers Agreed, but this is stated in the last paragraph and the example nicely illustrates the mode of failure and why it doesn't resolve the issue. I'm guessing that the answer to the OP's question is that there is no way to have a definition that always works. Also, looking at your profile, it sounds as though you may be able to give this answer definitively, so it would be a nice contribution. – Selene Routley Jun 20 '16 at 02:04
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@WetSavannaAnimalakaRodVance Yes, it had a lack of rigor. I just didn't want to get too technical to make it intituive. – Jun 20 '16 at 02:06
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2@Wet My answer would be identical to Zachary Selk's comment in the original question. – Jun 20 '16 at 02:18
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@T.Bongers Yes, but can you say why no definition will work. This answer shows how one attempted "obvious" definition won't work; can one give some more general insights into why other attempted definitions won't or even can't work. – Selene Routley Jun 20 '16 at 02:28
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@T.Bongers Regarding the usefulness of this answer: If I have understood correctly, this answer seems very useful for getting around a technical issue from one of my other questions where I try to construct an $L^2$ function using $f(0)$ http://math.stackexchange.com/questions/1597834/relationship-between-decay-of-fourier-transform-and-smoothness-in-l2
Instead of $f(0)$ I can say, pick a representative of $f$ and then also choose one of its Lebesgue points $p$ and use $f(p)$
Whereas the comment to my original post is not so informative.
– Mark Jun 20 '16 at 03:17
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The problem here is that $\{0\}$ is a set of measure 0 in $\mathbb{R}$, and in general the restrictions to sets of measure 0 are not well defined. But one can say the following. If $M^{1/2}$ a subset of $\mathbb{R}^n$ with holder exponent 1/2 boundary, then the 'restriction' operator, defined a priori for functions in $C^\infty(M^{1/2})$, is a continuous map from the sobolev space $H^{1/2}(M^{1/2}) \to L^2(\partial M^{1/2})$ and hence can be extended to all of $H^{1/2}(M^{1/2})$. Thus when we assume that the functions has half a derivative of regularity, one can define the restriction of a function to $\partial M^{1/2}$ - a set of measure 0 in $M^{1/2}$. The keyword here is the trace theorem if you want to read more. I highly recommend Prof. Pierre Germain's notes, or Lions and Magness.
Hari Rau-Murthy
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I take it by "Hölder exponent 1/2 boundary$ you mean the Hölder continuous image of e.g. the unit circle? – Selene Routley Jun 20 '16 at 02:32
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1I mean that if you are given an atlas, i.e. have an open covering $\cup U_\alpha \supset \partial M^{1/2}$, and you have homeomorphisms $\phi_\alpha: R^n_+ \to U_\alpha$ then $\phi_{\alpha_2}^{-1} \circ \phi_{\alpha_1}|_{\mathbb{R^{n-1}}}$ is going to be holder continuous of exponent 1/2. – Hari Rau-Murthy Jun 20 '16 at 02:45
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Thank you for this answer. It seems to be a little advanced for my current level maybe it will be useful for others here – Mark Jun 20 '16 at 03:19
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wait, I think I know you ! Did you go to Courant and sit in the algebraic geometry seminar? – Hari Rau-Murthy Jun 20 '16 at 03:31
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