# 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

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.