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In the general situation of $f:S\to \mathbb R^m$ where $S\subset \mathbb R^n$. There is a form of the mean value theorem: $a\cdot (f(y)-f(x))=a\cdot (f'(z)(y-x))$ which requires a vector $a$ and dot products.

In Tom Apostol's Mathematical Analysis (Second Edition), page No. 355, I found that after choosing $a$ to be a unit vector and using Cauchy-Scwarz inequality, they have written $\parallel f(y)-f(x)\parallel\leq \parallel f'(z)\parallel \parallel y-x\parallel$. But how have they got rid of $a$ in the left hand side. If I choose the unit vector in the direction of the vector $f(y)-f(x)$, then it is possible, but how will it follow for an arbitrary unit vector? Please help!

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    $\begingroup$ You do exactly what you wrote. Take $a$ to point in the direction of $f(y) - f(x)$. On the right-hand side, use Cauchy Schwarz. $\endgroup$ – PhoemueX Aug 14 '15 at 16:58
  • $\begingroup$ That works...but my question is why is it not mentioned in Apostol? Is it a typos? $\endgroup$ – Anupam Aug 14 '15 at 17:00
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From $\,$C.Pugh $\,$ Real Mathematical Analysis $\,$ (2002) at the end of the proof of theorem 11, p. 277 $\,$(just the MVT), one reads

A vector whose dot product with every unit vector is no larger than $M\,|q-p|$ has norm $\le M\,|q-p|$ .

Probably Apostol refers to the same property, that is $$a \cdot z \le \lambda \quad \forall a:\;\|a\|=1 \quad\Rightarrow\quad \|z\|\le \lambda$$ one can prove by contradiction.

So $\,a\,$ is truly arbitrary. Perhaps the statement exists somewhere in the book $\,\dots$

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how will it follow for an arbitrary unit vector?

How what follows for an arbitrary unit vector? There is a claim that $$\| f(y)-f(x)\|\leq \| f'(z)\| \| y-x\| \tag{1}$$ for all in $x,y$. There are no unit vectors in the inequality $(1)$.

In order to prove (1), one introduces a unit vector $a$ in the direction of $f(y)-f(x)$, and applies a previously proved statement to these $x,y$, and $a$.

When you introduce an object into a proof, you get to decide what you want it to be. There is no need to consider arbitrary unit vectors here.

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  • $\begingroup$ This is pretty good. I have also done like this. But it is mentioned on Page No. 356 Example 2 in Mathematical Analysis by Apostol that if a is an unit vector then the claim will be proved. That is my only confusion. In that book it is never mentioned that a is an unit vector in the direction of f(y)-f(x). $\endgroup$ – Anupam Aug 16 '15 at 11:27
  • $\begingroup$ I suppose the author did not express himself well, but the above is what he meant. $\endgroup$ – user147263 Aug 16 '15 at 16:01

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