2
$\begingroup$

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.

$\endgroup$
  • 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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 $$\sum_{i=1}^n \frac{\partial f}{\partial x_i}(x^{*})\cdot(x_i-x'_i)=f(x)-f(x').$$

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
1
$\begingroup$

I think the sequence for the proof must be,

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.