I have some questions concerning a proof of the Mean Value Theorem from Bredon's book that I find rather cryptical.

Theorem: Let $f : \mathbb{R}^n \to \mathbb{R}$ be $C^1$. Let $x := (x_1, \dots , x_n)$ and let $x' := (x'_{1}, \dots , x'_n)$. Then there exists a $x^* := (x^*_{1}, \dots , x^*_n)$ on the line segment between $x$ and $x'$ such that $$ f(x) - f(x') = \sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) (x - x').$$

Proof: Apply the Mean Value Theorem found in any calculus book to the function $\mathbb{R} \to \mathbb{R}$ defined by $t \mapsto f(tx + (1-t)x')$ and use the Chain Rule: $$ \frac{d f(tx + (1-t)x')}{dt} \Bigg|_{t = t^*} = % \sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) \frac{d (tx_i + (1-t)x'_i)}{dt} \Bigg|_{t = t^*} = % \sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) (x_i - x'_i )$$ with $x^* := t^* x + (1 - t^* )x' $.

I have the following problems.

  1. Why $t \mapsto f(tx + (1-t)x')$ is defined on $\mathbb{R} \to \mathbb{R}$ ?
    I think it should rather be defined on $[0,1] \to \mathbb{R}$ (i.e., it should be convex and not affine), since we want to capture that $x^* \in f(t^* x + (1 - t^* )x')$ for some $t^* \in [0,1]$.

  2. If $t \mapsto f(tx + (1-t)x')$ is defined on $[0,1]$ then I can almost get what happens in the equation. That is, things should work as follows: start from $t \mapsto f(tx + (1-t)x')$ and apply calculus Mean Value Theorem getting \begin{align} \frac{f(x) - f(x')}{?} & = \frac{df}{dt} (t^*) \\ & = \sum^{n}_i \frac{\partial f}{\partial x_i} (x^*). \end{align} Notice that at the denominator on the LHS I put a question mark because I don't see how it should work. We should get $(x - x')$, but I don't really see how. Or better, I see it, if the denominator works as $$ (1x + (1-1)x' - 0x + (1-0)x' ),$$ which is indeed equal to $(x - x')$, but to me this is not clear at all. Indeed it should simply be equal to $1$, because we are acting on the domain of $t \mapsto f(tx + (1-t)x')$, that should be just $[0, 1]$, not on its codomain.
    [The RHS should be OK, since we have that there is a $t^*$ and then we use it to define $x^*$.]

How should this actually work?

Any feedback would be greatly appreciated since I am self-taught and I think I never really got how these analysis proofs with differentiation actually work.

Thanks a a lot for your time.

  • 2
    $\begingroup$ Main idea: You've got two points in $\mathbb{R}^n$. Draw the line between the points and restrict $f$ to that line. The line looks like $\mathbb{R}$, so you apply the standard MVT to the line and reinterpret. $\endgroup$ – Michael Burr Apr 11 '17 at 15:09
  • $\begingroup$ @MichaelBurr: First of all, thanks a lot for the reply! Thus, concerning my point (1) indeed $t \mapsto f(tx + (1-t)x')$ should be with $t \in [0,1]$, i.e., convex and not affine, right? $\endgroup$ – Kolmin Apr 11 '17 at 15:11
  • $\begingroup$ @MichaelBurr: Actually, I think I see the logic behind the proof. With this question I rather wanted to check my understanding of the calculations behind the proof (e.g. point (2), which really leaves me puzzled). I am rather shaky on this aspect, in particular in this context... :-). $\endgroup$ – Kolmin Apr 11 '17 at 15:13
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    $\begingroup$ The definition in $(1)$ works for all of $\mathbb{R}$, there is no need to restrict it to $[0,1]$, but one could. It means working with a segment instead of a line ... $\endgroup$ – Michael Burr Apr 11 '17 at 15:21
  • $\begingroup$ @MichaelBurr: Again, thanks a lot! Actually I was kind of afraid of getting this feedback, because now I really don't see what happens in my point (2) (that is, I really don't see what is going on in the calculation in the proof). $\endgroup$ – Kolmin Apr 11 '17 at 15:25

In a sense, the MVT here is NOT applied to $f(x)$, at least not directly. You have several functions here:

  1. $f(x):\mathbb{R}^n\to\mathbb{R}$, the given one;
  2. $h(t):\mathbb{R}\to\mathbb{R}^n$ via $h(t)=tx+(1-t)x'$ for the fixed points $x$ and $x'$, describing the line segment connecting them;
  3. and their composition $g(t):\mathbb{R}\to\mathbb{R}$ via $g(t)=f(h(t))=f(tx+(1-t)x')$.

It's the last one to which we can apply the MVT on the interval $[0,1]$. According to the MVT (I'm skipping verification of its conditions), there exists a point $t^{*}\in(0,1)$ such that $$g'(t^{*})=\frac{g(1)-g(0)}{1-0}=g(1)-g(0).$$ The right-hand side is $$g(1)-g(0)=f(h(1))-f(h(0))=f(x)-f(x'),$$ i.e. the desired difference at the endpoints. For the left-hand side we apply the Multivariate Chain Rule to $g=f(h(t))$, where $h(t)=(h_i(t))_{i=1}^n=(tx_i+(1-t)x'_i)_{i=1}^n$: $$\frac{dg}{dt}=\sum_{i=1}^n \frac{\partial f}{\partial h_i}\cdot\frac{dh_i}{dt}=\sum_{i=1}^n \frac{\partial f}{\partial x_i}\cdot(x_i-x'_i).$$ So if we denote $x^{*}=h(t^{*})=(t^{*}x_i+(1-t^{*})x'_i)_{i=1}^n$ to be the point on this line segment with the "coordinate" $t^{*}$, we have the desired result that $$\sum_{i=1}^n \frac{\partial f}{\partial x_i}(x^{*})\cdot(x_i-x'_i)=f(x)-f(x').$$

  • $\begingroup$ Thanks a lot for this very very nice answer, which is exactly what I was looking for! Just one question, which is related with MichaelBurr's comments below the question. As I do, you assume that $t \in (0, 1)$. How does the entire thing works when $t \in \mathbb{R}$ as MichealBurr was suggesting is possible? $\endgroup$ – Kolmin Apr 11 '17 at 16:35
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    $\begingroup$ Indeed it is defined for $t\in\mathbb R$. The fact is the proof necessitates $t\in[0,1]$ and it's indifferent about what happens out from there. $\endgroup$ – Rafa Budría Apr 11 '17 at 16:42
  • $\begingroup$ @RafaBudría: Thanks a lot for the comment! Hence, in a sense what I wrote in point (1) did make sense. I wonder why not being explicit by just setting $t \in [0,1]$, but that's the choice of Bredon. :-) $\endgroup$ – Kolmin Apr 11 '17 at 17:39

I think the sequence for the proof must be,

$$\frac{d f(tx + (1-t)x')}{dt} \Bigg|_{t = t^*} = % \sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) \frac{d (tx_i + (1-t)x'_i)}{dt} \Bigg|_{t = t^*} = % \sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) (x_i - x'_i ) $$

  • $\begingroup$ Thanks a lot for having spotted this typo. However, the question is actually beyond it. $\endgroup$ – Kolmin Apr 11 '17 at 16:32
  • $\begingroup$ It seemed to me that the typo obscured all the proof. $\endgroup$ – Rafa Budría Apr 11 '17 at 16:38

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