# Given a real valued C1 function, show there exists a continuous multivalued function F with f(x) = x dot F(x)

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 at 21:41 Yes, I'll fix that. – Freddie Jan 9 at 21:59

## 1 Answer

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 at 9:32