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?
[1] Jittorntrum (1978) - TECHNICAL NOTE: An Implicit Function Theorem
 A: Too long for a comment.

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 $h(U)\ni h(x_2)$.
A: 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.
