Does being in C2 imply being in L2 for a function?

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Discussion Overview

The discussion revolves around whether a function that is twice continuously differentiable on the boundary of a bounded set automatically belongs to the space of square-integrable functions on that boundary. The scope includes theoretical implications of function spaces and properties of differentiability.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant questions if a function ##g\in C^2(\partial U)## implies ##g\in L^2(\partial U)##, seeking a simple yes or no answer.
  • Another participant asserts that the answer is no, providing an example related to the function ##f(x)=\frac{\sin (x)}{x}##, but suggests that if ##\partial U## is compact, the answer might be yes.
  • A third participant agrees with the previous assertion but notes that the boundedness of the domain ##U## could make the initial example invalid, implying that the conditions on the boundary may ensure continuity on the closure of ##U##.
  • A later reply acknowledges the boundedness of ##U## and concludes that the result is true under these conditions.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between ##C^2## and ##L^2##, with some suggesting that boundedness of the domain may lead to agreement on the implications, while others maintain that it does not automatically follow.

Contextual Notes

There are unresolved assumptions regarding the nature of the boundary and the implications of compactness, as well as the specific conditions under which the relationship between ##C^2## and ##L^2## holds.

TaPaKaH
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Having bounded set ##U\subset\mathbb{R}^n## with ##C^1## boundary ##\partial U## and a function ##g\in C^2(\partial U)##, does one automatically have ##g\in L^2(\partial U)##?
I don't need a proof/explanation, yes/no answer is sufficient.
 
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I assume ##C^2## is twice continuously differentiable and that ##L^2## square (Lebesgue)-integrable.

The answer is no. An example is given by some modification of the function ##f(x)=\frac{\sin (x)}{x}##.

If ##\partial U## is compact, then the answer is probably yes.
 
micromass said:
I assume ##C^2## is twice continuously differentiable and that ##L^2## square (Lebesgue)-integrable.

The answer is no. An example is given by some modification of the function ##f(x)=\frac{\sin (x)}{x}##.

If ##\partial U## is compact, then the answer is probably yes.

Since the domain (U) is bounded, then your example is bad. U bounded and the condition on the boundary may be sufficient to insure continuity on the closure of U, which is compact.
 
I somehow missed that ##U## was supposed to be bounded. The result is true then.
 

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