# Showing that $g(T)\subseteq f'(I) \subseteq\overline{g(T)}$ where $g(x,y) = \frac{f(x) - f(y)}{x-y}$

I'm trying to solve a three part exercise.

The first part asks me to show that the set $$T=\{(x,y)\in I\times I : x < y\}$$, where $$I$$ is just some open interval in $$\mathbb{R}$$ is connected.

I've done this by showing that it's path connected (just take a straight line path between any two points)

The second part - the one I'm having trouble with - asks me to show that, for a differentiable function $$f:I\rightarrow\mathbb{R}$$ and for $$g: T \rightarrow \mathbb{R}$$ defined by

$$g(x,y) = \frac{f(x) - f(y)}{x-y}$$

that we have $$g(T)\subseteq f'(I) \subseteq\overline{g(T)}$$, where $$\overline{g(T)}$$ is just the closure of $$g(T)$$.

I can see how to solve the third part which comes after this provided I can do this part, but I'm really struggling to see how to show these inclusions.

Any help you could offer would be really appreciated.

• Can you see that $$g(T) \subseteq f'(I)$$ is an obvious consequence of Lagrange's theorem? That's because $$\frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c \in (x;y)$. The other inclusion $f'(I) \subseteq \overline{g(T)}$ follows from the definition of derivative. – Crostul Feb 6 at 14:10

## 1 Answer

The first inclusion, as stated in the comments, is a consequence of Lagrange's theorem. As for $$f'(I) \subseteq \overline{g(T)}$$, a possible argument would be the following. Any point in $$\overline{g(T)}$$ is the limit of a $$g(T)$$-valued sequence. Now, consider some $$z \in f'(I)$$, so $$z = f'(x)$$ for some $$x \in I$$. This implies that the following equality holds (in particular, the following limit exists), $$$$\lim_{\varepsilon \rightarrow 0}\frac{f(x+\varepsilon)-f(x)}{\varepsilon} = f'(x).$$$$ This implies that the sequential limit exists, and in particular that the sequential left limit exists, so that for some $$I$$-valued sequence $$x_n$$ converging to $$x$$ from the left (i.e. for every $$n$$, $$x_n < x$$) we have $$$$\lim_{n \rightarrow +\infty}\frac{f(x_n)-f(x)}{x_n - x} = f'(x).$$$$ Now, for any $$n$$ we have that $$(x_n, x) \in T$$ since $$x_n < x$$, so $$(x_n, x)$$ is a $$T$$-valued sequence. Call it $$r_n := (x_n, x)$$, and thus the equation turns into $$$$\lim_{n \rightarrow +\infty} g(r_n) = f'(x),$$$$ which shows that $$z$$ is the limit value for some $$g(T)$$-valued sequence and thus belongs to its closure.