- #1

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I don't need a proof/explanation, yes/no answer is sufficient.

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- Thread starter TaPaKaH
- Start date

- #1

- 54

- 0

I don't need a proof/explanation, yes/no answer is sufficient.

- #2

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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.

- #3

mathman

Science Advisor

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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.

- #4

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I somehow missed that ##U## was supposed to be bounded. The result is true then.

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