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

1. Apr 26, 2013

1. The problem statement, all variables and given/known data
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 $C^{2}$.

3. 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.

Last edited: Apr 26, 2013
2. Apr 28, 2013

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.

3. Apr 28, 2013