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I've got a question regarding the MVT for vector valued functions which states

$$ f(x)-f(y)= \int_0^1 Df(tx+(1-t)y) dt (x-y) $$ where $ f: \mathbb{R}^n \to \mathbb{R}^n $ differentiable with Jacobian $Df()$ and $x,y \in \mathbb{R}^n$, or in a convex compact set.

The MVT for $f: \mathbb{R}^n \to \mathbb{R} $ states $$ f(x+h)-f(x)=\nabla f(\psi)^Th $$ for some $\psi$ one the line x to x+h.

So if I derivate this term with respect to x(if f is twice differentiable) I got

$$\nabla f(x+h)-\nabla f(x)=\nabla^2 f(\psi)h $$

where $\nabla f:\mathbb{R}^n \to \mathbb{R}^n $

so am I right to assume there exisits an $\psi \in \mathbb{R}^n $ so that this equation holds? Why is that (not) the case? I dont quite understand why we need to integrate the whole term?

Thank you for your help. Greetings.

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Consider $f(x,y)=x^4+y^3.$ It is $\nabla f(0,0)=(0,0),\nabla f(1,1)=(4,3)$ and $$\nabla^2 f(x,y)=\begin{pmatrix} 12x^2&0\\0& 6y\end{pmatrix}.$$ Now, $$\nabla f(1,1)-\nabla f(0,0)=(4,3)$$ and $$\nabla^2 f (\theta,\theta)\cdot (1,1)^T=\begin{pmatrix}12\theta^2&0\\0& 6\theta\end{pmatrix}\begin{pmatrix}1\\ 1\end{pmatrix}=\begin{pmatrix}12\theta^2\\ 6\theta\end{pmatrix}.$$

But $$\begin{cases}12\theta^2&=4\\6\theta&=3\end{cases}\iff \begin{cases}\theta&=\pm\frac{1}{\sqrt{3}}\\\theta&=\frac{1}{2}\end{cases}$$ which is impossible.

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I add some comments. The vector phi in the mean-value theorem depent on x. Therefore, the differentiation is not correct, because phi become a function.

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