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

[TaPaKaH]

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.

 

[micromass]

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]

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.

 

[micromass]

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