This is a problem from a previous graduate preliminary exam in multivariable analysis/calculus that I am trying to solve for my own practice:


Let $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ be a twice continuously differentiable function satisfying: $f\left ( 0,y \right )=0$ for all $y\in \mathbb{R}$.

1- Prove that $f\left ( x,y \right )=x.g\left ( x,y \right )$ for all $\left ( x,y \right )\in \mathbb{R}^{2}$, where: $g\left ( x,y \right )=\int_{0}^{1}\frac{\partial f\left ( tx,y \right )}{\partial x}dt$.

2- Show that $g$ is continuously differentiable, and that for all $x$ in $\mathbb{R}$:

$g\left ( 0,y \right )=\frac{\partial f}{\partial x}\left ( 0,y \right )$ and $\frac{\partial g}{\partial y}\left ( 0,y \right )=\frac{\partial ^{2}f}{\partial x\partial y}\left ( 0,y \right )$

For the first part: The only thing I could do so far is the following: I am basically trying to start from the right hand side to reach the left hand side. Note that: $g\left ( x,y \right )=\int_{0}^{1}\frac{\partial f\left ( tx,y \right )}{\partial x}dt=\int_{0}^{1}t\frac{\partial f\left ( u,y \right )}{\partial u}dt$ where $u=tx$. Then, I have no idea how to go forward?

Any help is appreciated.


Your formula $$g\left ( x,y \right )=\int_{0}^{1}\frac{\partial f\left ( tx,y \right )}{\partial x}dt=\ldots$$ is not correct as written: ${\partial f(tx,y)\over\partial x}$ is not the same as $f_x( tx,y)$. That's why I prefer to write $f_{.1}$ instead of $f_x$ in cases where the first variable (originally $x$) becomes an expression containing $x$.

Here is a correct version of your argument: $$f(x,y)=f(0,y)+\int_0^x f_{.1}(x',y)\ dx' = x\int_0^1 f_{.1}(tx,y)\ dt\ .$$ This shows that $g$ can be written as $$g(x,y)=\int_0^1 f_{.1}(tx,y)\ dt\ ,$$ and since $f\in C^2$ it follows that $g\in C^1$.


For the first part, you are almost done. Observe that when $x = 0$, $xg(x,y) = 0 = f(x,y)$ (the right side follows by hypothesis). It remains to show that for all nonzero $x$, we have

$$ g(x,y) = \frac{1}{x}f(x,y), $$

which, by definition of $g$, should hold if and only if for all $x\neq 0$,

$$ f(x,y) = x\int_0^1\frac{\partial f(tx,y)}{\partial x}dt = \int_0^1 x\frac{\partial f(tx, y)}{\partial x}dt. $$

Now set $u = tx$. Then $du = xdt$; moreover, when $t = 1$, $u = x$. Hence the integral on the right hand side above becomes

$$ \int_0^x\frac{\partial f(u,y)}{\partial x}du, $$

which by fundamental theorem of calculus is equal to

$$ f(x,y) - f(0,y) = f(x,y) $$

(the equality above follows by hypothesis on $f$).

Now try to do the second part regarding properties of $g$.

  • $\begingroup$ Shouldn't the integral $$\int_0^x\frac{\partial f(u,y)}{\partial x}du$$ be equal to: $$\int_{0}^{x}\frac{\partial f\left ( u,y \right )}{\partial u}.\frac{\partial u}{\partial x}du=\int_{0}^{x}t\frac{\partial f\left ( u,y \right )}{\partial u}du$$ ? and then you apply the fundamental theorem of calculus, but the appearence of $t$ is creating a problem now and the theorem cannot be applied because of that $\endgroup$
    – M.Krov
    Feb 13 '12 at 7:47
  • $\begingroup$ Because it should be $\partial_x f(tx,y)$ and not $\partial f(tx,y)/\partial x$ (i.e. one has to differentiate after changing the variable $x\mapsto tx$ and not before (see also Christian's answer above). $\endgroup$
    – user2093
    Feb 14 '12 at 20:22

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