I want to prove $$f(x) - f(0) = \sum_{j=1}^nx_j\int_0^1 \frac{\partial}{\partial x_j} f((1-s)x) ds$$ for $f\in \mathcal{C}^\infty,$ and $x\in \mathbb{R}^n$. To be clear, $f: \mathbb{R}^n\to \mathbb{R}.$ I have tried various things but none have worked. Any thoughts?
Evaluating this integral/sum
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real-analysis
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0Fix $x$, and consider $g \colon \mathbb{R} \to \mathbb{R}$ given by $g(t) = f(t\cdot x)$. FTC on $g(1) - g(0)$ plus a substitution. – 2017-02-11
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0I'm afraid I don't quite understand your hint. Could you elaborate a bit more? – 2017-02-11
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0What is $g'(t)$? – 2017-02-11
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0If $g(t) = f(t\cdot x)$ then $g'(t) = xf'(t\cdot x)$. Are you saying I should apply the FTC where $F(x) = \int_a^x f(t) dt$ or the FTC where $\int_0^1 f(t) dt = F(1) - F(0)$? I have tried variations of the second and I can't get it to work. – 2017-02-11
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0If $f \colon \mathbb{R}^n \to \mathbb{R}$, what do you mean with $f'$? – 2017-02-11
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0Good question. I guess I don't actually know. – 2017-02-11
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0So what does the multivariable chain rule say is $g'(t)$? – 2017-02-11
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0Ah, I think I see the issue. I forgot that order is really rather important: $g'(t)$ should be $f'(t\cdot x) \cdot x$, no? – 2017-02-11
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0Maybe. Depends on what you mean by that. – 2017-02-11
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0Perhaps I am being dense and not seeing the point of your hint, or missing something very obvious. – 2017-02-11
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0I mean that it is $(\frac{\partial f}{\partial x_1} , \dotsc, \frac{\partial f}{\partial x_n}) (t\cdot x) \cdot x$ where $x = (x_1, \dotsc, x_n)$. – 2017-02-11
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0Ah hah. And if you write out the dot product, does that start to look like it's related to your sum of integrals? – 2017-02-11
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0Ah, I think I see now. I can make it look like the sum of integrals without the integral sign; integrating $g'(t)$ should give the desired result. Thank you! – 2017-02-11