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Consider the following theorem:

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It appears in many different places but always with both conditions stated:

One: $g_i(0,...,0) = {\partial f\over \partial x_i}(0,...,0)$

Two: $f(x_1,...,x_n) = \sum_i x_i g_i$

But as far as I can tell the first one follows from the second one. Consider the case in two dimensions: Assume $f(x,y) = xg(x,y) + yh(x,y)$ where $f,g,h$ are smooth. Then $$ {\partial f \over \partial x } (x,y) = g(x,y) + x {\partial g \over \partial x } (x,y) + y {\partial h \over \partial x } (x,y)$$

and then

$${\partial f \over \partial x } (0,0) = g(0,0)$$

The same argument works for $n$ dimensions. What am I missing here? (Surely if one really did follow from two then only two would be stated?)

If "two" then of course the $g_i$ are already shown to exist.

newb
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1 Answers1

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HINT: This is also called Hadamard's lemma. One of the most simple proofs involves connecting the origin and $x$ with a straight line and representing the restriction of $f$ on that line via fundamental theorem of calculus (just because restriction is a function on the segment in $\mathbb{R}$)

Evgeny
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  • Sorry, could you say a bit more? I don't understand yet. – newb Oct 02 '13 at 12:24
  • Well, it seems that only part that confuses you is simultaneous presence of these two sentences.You're totally right that if function has such representation, then values of $g_i$'s at origin are $\frac{\partial f}{\partial x_i} (\overline{0})$. Rigorously it can be proven in slightly another way: if $f$ is $C^{r}$, then $g_i$'s are $C^{r-1}$ and if $f$ was $C^1$ we cannot use Leibniz rule. But modification of proof is very easy. My answer was addressed to case if you wanted to find proof for this lemma. – Evgeny Oct 02 '13 at 15:31
  • Thank you. So this really is a theorem of the form "If $A$ then $B$ and $C$" where $B \implies C$, huh. Weird. – newb Oct 12 '13 at 10:22
  • @newb The formulation of this theorem was a statement of this form :) I think, it was formulated such way to make explicit both statements – Evgeny Oct 12 '13 at 12:28