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

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.

## Answers and Replies

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.

mathman
Science Advisor
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.