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Assume $f:\mathbb{R}^{n} \rightarrow \mathbb{R}$ is a function with continuous first order partial derivatives such that $f(0)=0$. Show there exists a continuous function $F:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ such that $f(X)=X \cdot F(X)$ on $\mathbb{R}^{n}$.

It seems like the function $F(X):=(\int_{0}^{1} (\partial_{j}f)(tX)dt))_{1\leq j \leq n}$ is the right idea, but it doesn't seem to work out. I think I'm missing something...would appreciate any help.

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You write $f(X) = X \cdot F(x)$; should these $X$ really be $x$ (should they be the same)? –  Muphrid Jan 9 '13 at 21:41
    
Yes, I'll fix that. –  Freddie Jan 9 '13 at 21:59

3 Answers 3

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It seems like the function $F(X):=(\int_{0}^{1} (\partial_{j}f)(tX)dt))_{1\leq j \leq n}$ is the right idea, but it doesn't seem to work out.

You are on the perfectly right track, but missed the target slightly. First notice the following identity about the partial derivative with respect to $t$: $$ \int^1_0 \frac{\partial f(tX)}{\partial t}dt = f(X) - f(0) = f(X).\tag{1} $$ Hence $$ \int^1_0 \frac{\partial f(tX)}{\partial t}dt = \int^1_0 X\cdot \nabla f(tX) \,dt = X\cdot \underbrace{\int^1_0 \nabla f(tX) \,dt}_{\text{This is } F(X)}. $$ This is the same with the expression for $F$ you wrote: $$ F(X):=\left(\int_{0}^{1} (\partial_{j}f)(tX)dt)\right)_{1\leq j \leq n}. $$


Remark: we don't necessarily need $f(0)=0$. For (1) can be modified to: $$ \int^1_0 \frac{\partial \big(t^n f(tX)\big)}{\partial t}dt = 1^n f(X) - 0^n f(0) = f(X). $$ The left side is $$ \int^1_0 \frac{\partial \big(t^n f(tX)\big)}{\partial t}dt = \int^1_0 \Big(nt^{n-1} f(tX) + t^n X\cdot \nabla f(tX)\Big)dt \\ = \nabla \cdot \int^1_0 t^{n-1} X f(tX)\,dt. $$

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For given $x$ consider the auxiliary function $\phi(t):=f(tx)$. By the chain rule one has $$\phi'(t)=\nabla f(tx)\cdot x$$ and therefore $$f(x)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\ dt=\left(\int_0^1 \nabla f(tx)\ dt\right)\cdot x\ .$$ It follows that $$F(x):=\int_0^1 \nabla f(tx)\ dt$$ does the job, as you have conjectured.

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Initially I thought $$ F(x)=\frac{xf(x)}{x.x}$$ would work, but sadly as pointed out is not defined at the origin. We could add the case $F(0)=0$ but this does not lead to a continuous definition unless we have the additional contraint that $\nabla f=0$ at the origin.

Fortunately we can construct such a function as follows:-

Let $H=\nabla f$ and $h=\frac{H(0)}n$

Now take $g(x)=f(x)-x.h$

Then $\nabla g = \nabla(f - x.h)=\nabla f-\nabla (x.h)=\nabla f-h\nabla.x-x\nabla.h$

$\nabla.x=n$ and as $h$ is constant $\nabla.h=0$

S0 $\nabla g=\nabla f-hn=\nabla f-H(0)$

And thus $\nabla g(0) = \nabla f(0) - H(0)=H(0)-H(0)=0$

So we can use my initial thought for $g(x)$ and take $$G(x)=\frac{xg(x)}{x.x}$$ with $G(0)=0$ which thanks to $\nabla g(0)=0$ is continuous everywhere, and $$ x.G(x) = x.\frac{xg(x)}{x.x}= \frac{x.xg(x)}{x.x}=g(x) $$ for $x\neq 0$ and $g(0)=f(0)-0.h=0-0=0.0$ also works at the origin, Hurrah!

But now we need to find something which will work for $f(x)$

Now $f(x)=g(x)+x.h=x.G(x)+x.h=x.(G(x)+h)$

so we can take $F(x)=G(x)+h$

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Very true, I've had to expand my answer a lot to fix that problem! –  Shard Jan 10 '13 at 9:32

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