Continuity in the Mean Value Theorem and the Implicit Function Theorem

Let $$f:\mathbb{R}^n\to\mathbb{R}^n$$ be a $$\mathcal{C}^2$$ function of $$x\in\mathbb{R}^n$$. Then we can use the mean value theorem (MVT) to formulate the following equality: $$f(x_1)-f(x_2)=\frac{\partial f}{\partial x}(x_2+\lambda(x_1-x_2))(x_1-x_2),$$ with $$\lambda\in(0,1)$$. This equality is point-wise true by the MVT for every pair $$(x_1,x_2)$$. Furthermore, it is easy to show with a simple example that for a pair $$(x_1,x_2)$$, there might be multiple $$\lambda$$ that yield the equality.

I'm currently looking in the continuation of $$\lambda$$ when $$x_1,x_2$$ vary continuously. There are also quite simple examples to formulate that show that continuous variation of $$x_1,x_2$$ yield discontinuous continuations of $$\lambda$$. (Check e.g. this example, and move point B to 4)

Now I want to show that for every pair $$(x_1,x_2)$$ you can find a neighborhood $$\mathfrak{B}$$ around $$(x_1,x_2)$$ for which there exists a continuous function $$\lambda(x_1,x_2)$$ such that for all $$(x_1,x_2)\in\mathfrak{B}$$, we have that $$f(x_1)-f(x_2)=\frac{\partial f}{\partial x}(x_2+\lambda(x_1,x_2)(x_1-x_2))(x_1-x_2)$$ and for any continuous trajectory that takes values in $$\mathfrak{B}$$, the function $$\lambda(x_1,x_2)$$ is continuous. Trying to counterexample this with multiple examples in geoabgebra (and others) did not help. Up until now I found that this must be true in my setting, I just don't know how to prove this...

What I tried: I tried to show this with the implicit value theorem by analyzing the implicit function $$\mathscr{F}(x_1, x_2, \lambda)= f(x_1)-f(x_2)-\frac{\partial f}{\partial x}(x_2+\lambda(x_1-x_2))(x_1-x_2),$$ However, the classic version requires a differentiability condition: $$\frac{\partial^2 f}{\partial x^2}\neq 0$$ at the point where $$\mathscr{F}=0$$, which I cannot guarantee perse - plus this seems like a very restrictive assumption for my case. I found a variation of the theorem in [1]. But this theorem requires a one-to-one relationship, which is for me not the case, as there might be multiple lambda's. This was the closed I could get, but no concrete results yet...

Can anybody help me with this problem?

• What is $\partial f/\partial x$ supposed to mean for a vector $x$? Jan 18, 2021 at 16:37
• The Jacobian, thus $\begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \cdots & \frac{\partial f_n}{\partial x_n}\end{bmatrix}$ Jan 19, 2021 at 9:11

However, the classic version requires a differentiability condition: $$\frac{\partial^2 f}{\partial x^2}\neq 0$$ at the point where $$\mathscr{F}=0$$, which I cannot guarantee perse - plus this seems like a very restrictive assumption for my case. I found a variation of the theorem in [1]. But this theorem requires a one-to-one relationship, which is for me not the case, as there might be multiple lambda's.
Unfortunately, some one-to-one conditions are necessary for the continuity of $$\lambda$$. For instance, suppose that $$n=1$$ and $$f’(x_2)\ne \frac{f(x_1)-f(x_2)}{x_1-x_2}$$. For each $$x$$ such that $$\lambda(x)$$ is defined put $$h(x)=x+\lambda(x_1,x)(x_1-x)$$. Then $$f’(h(x))=\frac{f(x_1)-f(x)}{x_1-x}$$ is monotonic (and so one-to-one) on some open neighborhood $$U$$ of $$x_2$$. Since $$h$$ is a continuous one-to-one map on a non-empty open subset $$U$$ of $$\Bbb R$$, by the invariance of domain $$h(U)$$ is an open subset of $$\Bbb R$$ and $$h$$ is a homeomorphism between $$U$$ and $$h(U)$$. Moreover, $$f’$$ is one-to-one on $$h(U)\ni h(x_2)$$.
• @seaver The value $\lambda(x_1,x_2)$ can be non-unique, but when we fix it, we fix $h(x_2)$ and conclude that $f’$ is one-to-one on $h(U)\ni h(x_2)$. Jan 22, 2021 at 5:48
I know you are not too enthusiastic about counterexamples, but there is a problem with your MVT. Explicitly, there may be no such $$\lambda(x_1,x_2)$$.
A simple example: Let $$n=2$$ and set $$f(x,y)=(x^2,y^3)$$. Obviously, $$f$$ satisfies your assumptions and has Jacobi matrix $$\frac{\partial f}{\partial(x,y)} = \begin{pmatrix}2x & 0 \\0 & 3y^2\end{pmatrix}.$$ Now for $$(x_1,y_1)=(0,0)$$ and $$(x_2,y_2)=(1,1)$$ we are looking for a solution of $$f(1,1)-f(0,0)=\begin{pmatrix}1\\1\end{pmatrix} \overset{?}{=} \begin{pmatrix}2\lambda & 0\\3\lambda^2 & 0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} =\begin{pmatrix}2\lambda\\3\lambda^2\end{pmatrix},$$ which can easily be seen does not exist.