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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Crostul Feb 6 at 14:10

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), \begin{equation} \lim_{\varepsilon \rightarrow 0}\frac{f(x+\varepsilon)-f(x)}{\varepsilon} = f'(x). \end{equation} 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 \begin{equation} \lim_{n \rightarrow +\infty}\frac{f(x_n)-f(x)}{x_n - x} = f'(x). \end{equation} 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 \begin{equation} \lim_{n \rightarrow +\infty} g(r_n) = f'(x), \end{equation} which shows that $z$ is the limit value for some $g(T)$-valued sequence and thus belongs to its closure.


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