Prove that a function defined by an integral is C^2

[DeadOriginal]

Homework Statement

Let n\geq 2 and define f:\mathbb{R}^{n}\rightarrow\mathbb{R} such that f is C^{2}. Suppose that there is a bounded S\subseteq\mathbb{R}^{n} such that f\restriction(\mathbb{R}^{n}\backslash S)=0. Define g:\mathbb{R}^{n}\rightarrow\mathbb{R} by g(u)=\int f(u+v)\log|v|dv. Show that g is C^{2}.

The Attempt at a Solution

Since f is C^{2} I was thinking that I could define a new function $$\varphi(u,v)=u+v$$ where \varphi:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}. Then f(u+v)=f\circ\varphi. Since the integral defined by g is integrated in terms of v we could just differentiate each term of g like so:
$$
\begin{align*}
\partial_{u_{1}}g(u)
&=\partial_{u_{1}}\int f\circ\varphi\log|v|dv=\int\partial_{u_{1}}(f\circ\varphi\log|v|)dv\\
&=\int(\partial_{u_{1}}f\circ\varphi)(\partial_{u_{1}}\varphi)\log|v|dv
\end{align*}
$$
and we would do this for all n elements of u. To show that g is C^{2} we could just differentiate twice. Of course I would also have to show that these partials are continuous.

Does this look correct or am I missing something big? Any help would be appreciated.

 

[christoff]

I'm assuming this is a problem for a course in real analysis? Then I think you're on the right track. However, you should first show, by referring to results you [should] already have seen, that
1) The function g actually exists (that is, the integral actually makes sense),
2) You actually CAN permute the derivative and integral like that.
The second of the two uses a fairly standard result (sometimes called the Leibniz integral rule).
After you've shown this, then your argument works fine.

 

[DeadOriginal]

Yes. This is for real analysis. Thanks for the heads up that I have to show that the function g actually exists. Otherwise I would have completely left that part out.