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I recently ran into the following exercise:

If $R(x)$ is a $C^\infty$-function near the origin in $\mathbb{R}^n$, satisfying $R(0)=0$ and $DR(0)=0$, show that there exist smooth functions $r_{jk}(x)$ such that$$R(x)=\sum r_{jk}(x)x_jx_k.$$

I know that the fundamental theorem of calculus applied to $\varphi(t)=F(x+ty)$ yields$$F(x+y)=F(x)+\int_0^1DF(x+ty)y\text{ }dt,$$provided that $F$ is $C^1$. Using this result, we can write $R(x)=\Phi(x)x$, $\Phi(x)=\int_0^1DR(tx)\text{ }dt$, since $R(0)=0$. Then $\Phi(0)=DR(0)=0$, so we can apply it again to obtain $\Phi(x)=\Psi(x)x$.

I get stuck here, however. Do you guys have any ideas on how to begin to prove that such smooth functions exist? Thanks in advance.

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1 Answer 1

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You are on the right track; but there is more linear algebra involved. Let $x\in{\mathbb R}^n$ be fixed and consider the auxiliary function $$\phi(t)\ :=\ f(t\, x)\qquad(0\leq t\leq 1)\ .$$ Then $$f(x)=\phi(1)=\phi(0)+\int_0^1 \phi'(t)\, 1\ dt =\phi(0)+\phi'(t)(t-1)\Bigr|_0^1\ +\int_0^1 \phi''(t)\,(1-t)\ dt\ .$$ Now $\phi(0)=f(0)=0$, then $$\phi'(t)=\sum_i x_i\ f_{.i}(t\,x) \ ;$$ in particular $\phi'(0)=0$ by assumption on $f$, and finally $$\phi''(t)=\sum_i x_i{\partial\over\partial t}f_{.i}(t\,x) =\sum_{i\,k} x_i\,x_k\, f_{.ik}(t\, x)\ .$$ It follows that $$f(x)=\sum_{ik}x_i\,x_k\ \int_0^1 f_{.ik}(t\, x)(1-t)\ dt\ ;$$ and $$r_{i\ k}(x)\ :=\ \int_0^1 f_{.ik}(t\, x)(1-t)\ dt$$ has the stated properties.

(Of course one could perform this argument coordinate-free, when desired.)

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