This is a text in my book:
The problem is that I have never seen a mean value theorem for vector-valued functions. I have seen it for functions of several varaibles, but then it must be a real function, not a vector function.
I have two questions:
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Is there any way to see their statement follow easy from a mean-value theorem?
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I wanted to prove their statement using only the mean value theorem for functions of several varaibles, is this proof correct?:
We have as in the text the function $\textbf{f}(\textbf{x},t)$.
For each component we have that $|\textbf{f}_i(\textbf{x}_1,t)|-\textbf{f}_i(\textbf{x}_2,t)|\le |\nabla_M|*|\textbf{x}_1-\textbf{x}_2|$. When they say continuous differentiable components, it means that their partial derivatives are continuous, hence the gradient has a maximum value $|\nabla_M|$ on the closed set?
And now since:
$\sqrt{a^2+b^2…+k^2}\le |a|+|b|…+|k|$, we get that
$|\textbf{f}(\textbf{x}_1,t)-\textbf{f}(\textbf{x}_2,t)|\le N*|\nabla_M|*|\textbf{x}_1-\textbf{x}_2|$, where N is the number of components.
Is this proof correct?
It became 3 questions, because I would also very much like someone to confirm what I wrote in bold text in question 2?
PS: I also actually used cauchy schwarz like this:
$|\textbf{f}_i(\textbf{x}_1,t)|-\textbf{f}_i(\textbf{x}_2,t)|=\nabla_{f_i}(c)\cdot|\textbf{x}_1-\textbf{x}_2|\le_{CS}|\nabla_{f_i}(c)|*|\textbf{x}_1-\textbf{x}_2|\le |\nabla_M|*|\textbf{x}_1-\textbf{x}_2|$
I used * for multiplying and $\cdot$ for dot-product.
Best Answer
Yes, your proof is correct. This is a very short answer, but there is little more I can add to your work.