not sure why this is smooth

**slevvio** · October 11th 2011, 09:17 AM

Define $g: V \rightarrow \mathbb{R}$ smooth where $V \subseteq \mathbb{R}^n$ is open. Then define a map, for a fixed $x \in V$

$h_i(x) = \int_0^1 \frac{\partial g}{\partial x_i}(tx) dt$

(we may assume tx is in V)

Can anyone explain to me why this function is smooth?

Thanks for any help

**xxp9** · October 11th 2011, 10:04 AM

$\frac{\partial{h_i}}{\partial{x^j}} = \int_0^1 t\frac{\partial^2{g}}{\partial{x^ix^j}}(t \mathbf{x}) dt$

**slevvio** · October 11th 2011, 02:22 PM

hmmm why is this the case?

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