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Hy everybody got a quick question. I know that all function F in a Sobolev Space has a continuous representative called U such as U=F almost everywhere.

Lets take for example: The Sobolev space on ]-1,1[

  • the function F(x)=0 on x<0 and F(x)=1 x>0 we can't find a continuous representative because it would be on a larger interval let say ]-eps ,eps[ so not in a Sobolev space

  • the function F(x)=x on x<0 and x>0 and F(0)=1 where we could find the representative U=x

My question is: if F isn't continuous is it sure that it isn't an element of a Sobolev Space? It was given was given as a necessary condition in my course.

Thanks for the help.

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  • $\begingroup$ Apparently you are thinking about a family of Sobolev spaces, which can mean changing the degree of differentiability or changing the domain (including perhaps dimension) on which functions are defined a.e. By including the specific definition of "Sobolev space" you will make answering your Question more tractable. $\endgroup$ – hardmath Nov 16 '13 at 17:57
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    $\begingroup$ I am speaking of H1 = W1,2 in one dimension $\endgroup$ – Htx12 Nov 16 '13 at 18:19
  • $\begingroup$ What you said you know at the beginning of your Question is correct, all functions in $H^1(-1,1)$ have a continuous representative. Functions in this space are technically only defined up to equality "almost everywhere", so you can change the values of a continuous U function on $[-1,1]$ arbitrarily on a set of measure zero, introducing if you wish infinitely many discontinuities as an equivalent representative F in this Sobolev space. Perhaps you are interested in a proof of the existence of the continuous representative U? $\endgroup$ – hardmath Nov 16 '13 at 19:14
  • $\begingroup$ Ok thanks I got the proof of the existence just wanted to be sure about about adding discontinuities on a set of measure zero would work. That's the answer i was looking for :) $\endgroup$ – Htx12 Nov 16 '13 at 21:05
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Your reasoning is correct. If $f$ cannot be redefined on a set of measure zero to become a continuous function, then $f$ is not in $H^1(-1,1)$.

Chances are that sooner or later you will encounter Sobolev functions of more than one variable. Those need not have a continuous representative. For example, $$ f(x,y) := \sqrt{- \log(x^2+y^2)} $$ is in $H^1(D)$ where $D$ is the disk in two dimensions. It cannot be made continuous, or even bounded, by changing its values on a set of measure zero.

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