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

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]n\geq 2[/itex] and define [itex]f:\mathbb{R}^{n}\rightarrow\mathbb{R}[/itex] such that [itex]f[/itex] is [itex]C^{2}[/itex]. Suppose that there is a bounded [itex]S\subseteq\mathbb{R}^{n}[/itex] such that [itex]f\restriction(\mathbb{R}^{n}\backslash S)=0[/itex]. Define [itex]g:\mathbb{R}^{n}\rightarrow\mathbb{R}[/itex] by [itex]g(u)=\int f(u+v)\log|v|dv[/itex]. Show that [itex]g[/itex] is [itex]C^{2}[/itex].


    3. The attempt at a solution
    Since [itex]f[/itex] is [itex]C^{2}[/itex] I was thinking that I could define a new function $$\varphi(u,v)=u+v$$ where [itex]\varphi:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}[/itex]. Then [itex]f(u+v)=f\circ\varphi[/itex]. Since the integral defined by [itex]g[/itex] is integrated in terms of [itex]v[/itex] we could just differentiate each term of [itex]g[/itex] 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 [itex]n[/itex] elements of [itex]u[/itex]. To show that [itex]g[/itex] is [itex]C^{2}[/itex] 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. jcsd
  3. Apr 28, 2013 #2
    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.
     
  4. Apr 28, 2013 #3
    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.
     
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