# Why is this proof incorrect? - $f(b) - f(a) = [Df(c)](b-a) = \nabla f(c) \cdot (b-a)$

I'm currently a student in a Vector Calculus class, and received my exam back today. I plan to go to the professor's office hours, but I'd like to ask here (in case there's some blatantly obvious fault I'm missing). Not only do I not understand why I received no partial credit in this problem, but I don't understand why it's not completely correct. I'd really appreciate any insight. The problem statement is:

Let $$a,b \in \mathbb{R^n}$$ with $$a \neq b$$. The segment $$(a,b)$$ in $$\mathbb{R^n}$$ is defined by: $$(a,b) := \{x \in \mathbb{R^n} | ~x = (1 - t)a + tb,~ t \in (0,1) \}$$

Let $$f: \mathbb{R^n} \rightarrow \mathbb{R}$$ be differentiable on $$\mathbb{R^n}$$. Use the Mean Value Theorem from single-variable calculus to show that there exists $$c \in (a,b) \subset \mathbb{R^n}$$ such that: $$f(b) - f(a) = [Df(c)](b-a) = \nabla f(c) \cdot (b-a)$$

Here's my proof:

We know $$f$$ is differentiable on $$\mathbb{R^n}$$ (including on the interval $$(a,b)$$). Fix all elements in $$x$$ but one, call this element $$x_i$$ (that is, the $$i^{th}$$ element of $$x$$. Let $$a_i < c_i < b_i$$. By the mean value theorem, we know there is at least one $$c_i$$ such that the tangent at $$c_i$$ is parallel to the secant between $$a_i$$ and $$b_i$$. That is, there exists a $$c_i$$ such that: $$\frac{f(b_i) - f(a_i)}{b_i - a_i} = \frac{\partial f(x_i)}{\partial c_i}$$ Multiplying by $$(b_i - a_i)$$ on both sides, we have: $$f(b_i) - f(a_i) = \frac{\partial f(x_i)}{\partial c_i}(b_i - a_i)$$ We now fix all other elements in $$x$$ and, by the same process, arrive at the same result for every component of $$x$$. This gives us: $$f(b) - f(a) = \left(\frac{\partial f(x_1)}{\partial c_1} + \frac{\partial f(x_2)}{\partial c_2} + ... + \frac{\partial f(x_n)}{\partial c_n} \right)(b-a)$$ Note that the term on the right side of the equality is just $$\nabla f(c) \cdot (b-a)$$, so we know that there exists some $$c$$ such that: $$f(b) - f(a) = \nabla f(c) \cdot (b-a)$$

I have pretty much zero idea where I went wrong (but it's, apparently, all wrong). I'm very confused.

• It is not clear what you mean by $f(a_i)$. $f$ is fed vectors, not scalars.
– Pedro
Mar 10, 2014 at 22:21
• How are you differentiating with respect to $c_i$? Mar 10, 2014 at 22:21
• You would need to show $c \in (a,b)$. (And this construction of $c$ won't get that result.) Mar 10, 2014 at 22:25
• @PedroTamaroff Once you fix all of the elements in the vectors but one, $f$ only varies with a single element—the scalar. I see where the differentiating w.r.t. thing is wrong; that's definitely a mess up on my part. I should've switched the $c_i$ and the $x_i$. It is assumed that $c \in (a,b)$, though; $c$ is just some point between $a$ and $b$. Mar 10, 2014 at 22:25
• @AmagicalFishy mmh are you using that $f(b) = \sum_i f(b_i)$? am I following your flow of thoughts right? Mar 10, 2014 at 22:28

mmh, if $x_i$ means the $i$-th coordinate of $x$, then you can't calculate $f(x)$ as $\sum_i f(x_i)$. Think about $f(x,y)=xy$. $f(1,1)=1$ but $f(0,1)=f(1,0)=0$. It seems (not so clear) that you are assuming some linear property over $f$.

If this is the case, your reasoning is ruined from the beginning. In fact the right approach to this exercise is to combine the function $f : \mathbb{R}^n \to \mathbb{R}$ with the function $h$ (the segment) $h: (0,1) \to \mathbb{R}^n$ to obtain a function $f(h(t)) :[0,1] \to \mathbb{R}$ for which the mean value theorem holds)

I'll add some hints for a correct resolution First of all, the mean value theorem holds for functions from $U \subseteq \mathbb{R} \to \mathbb{R}$. (it's very easy to see - even graphically - that even in 2 dimensions there are lots of counterexamples). So the first step is building such function. The idea is to observe $f$ in the direction of the segment $g: [0,1] \to \mathbb{R}^n \ g(t) = a+(b-a)t$, in this way, we can "consider" $f$ as a function from $\mathbb{R} \to \mathbb{R}$. The formalization of this reasoning is to make the composition of the two functions mentioned above to obtain $$g:= f(h(t)) : [0,1] \to \mathbb{R}$$. It's easy to see that $g$ is differentiable and you need to apply to it the mean value theorem. So the problem is, how to compute $g'$? Here comes in help the chain rule :)

• The domain of $h$ is $(0,1)$, not $\mathbb{R}$. Mar 10, 2014 at 22:34
• @aschepler yeah, fixing it now. thanks^^ (only want to stress the fact that the domain is contained in $\mathbb{R}$) Mar 10, 2014 at 22:35
• Ah! Damnit, what a dumb mistake. I was so proud of this "proof", too. Would you be able to elaborate on the appropriate way to approach this? I'm not sure I understand the process. Mar 10, 2014 at 22:39
• @AmagicalFishy Yeah, it's all right if I post the general fact and I leave to you the calculations? I think it's better for you:) so let me edit the answer :) Mar 10, 2014 at 22:41
• Absolutely. A general idea is fine. Thanks. :) Mar 10, 2014 at 22:42

You are writing $$\frac{f(b_i) - f(a_i)}{b_i - a_i} = \dots$$ What is that supposed to mean? $f$ is a function of $n$ variables. You are setting the $i$th variable equal to $a_i$ and to $b_i$. What about all the other variables? What are they equal to in this formula? Are they still variable?

I'm sorry, I think you're out of luck. No partial credit here.

• "Fix all elements in $x$ but one, call this element $x_i$ (that is, the $i^{th}$ element of x.)" Mar 10, 2014 at 22:41
• Show me how to do this for $f(x_1,x_2,x_3) = x_1x_2^2x_3^3$ and $a = (1,2,3)$, $b = (2,3,4)$. Mar 10, 2014 at 22:45
• I'm... not sure I understand what you mean. If we fix, say, $x_2$ and $x_3$, then $f(x_1,x_2,x_3) = f(x_1)$. Mar 10, 2014 at 22:48
• PLease write down a formula in this particular case. Mar 10, 2014 at 22:51
• A formula for what? I'm afraid I'm... not quite sure what kind of formula you're looking for. I think there's some misunderstanding; perhaps I'm not grasping what it really means to fix some variables? Mar 10, 2014 at 22:56