Finding $g_i:\mathbb{R}^n\to\mathbb R$ s.t $f(x)=\sum\limits_{i=1}^nx_i\cdot g_i(x)$

338 Views Asked by At

Let $f:\mathbb R^n\to\mathbb R$ differntiable and $f(0)=0$. Prove exist $g_i$ s.t for $x=(x_1,\dots,x_n):f(x)=\sum\limits_{i=1}^nx_i\cdot g_i(x)$. hint:$f(x)=\int\limits_0^1f\prime(tx)dt$.

I dont have any idea how to solve it. maybe we can build $g_i=f(x_i\cdot x)$ but im not sure it may help since I cannot take one time $\int f(x_1\cdot x)dx_1$ and in another $\int f(x_n\cdot x)dx_n$).

how can I build the sequence of functions and furthermore, what is the intuition for this type of questions involves building functions $\varphi:\mathbb R^n\to\mathbb R^m$ based on given condition with sums or diffrentiablity or partial derivatives?

1

There are 1 best solutions below

8
On BEST ANSWER

By the chain rule we shall have $$f'(tx)=\sum_{k=1}^n\frac{\partial f}{\partial x_k}(tx)x_k,$$ put $g_k(x):=\int_0^1 \frac{\partial f}{\partial x_k}(tx)dt.$ This is a form of Hadamard's lemma. I actually don't understand what you mean by your second question, which is very imprecise, but this lemma is used to prove, say, that every derivation of $C^\infty(\mathbf{R}^n)$ is a first order linear partial differential operator.