Corollary to Theorem 5.12 in Rudin's PMA: The derivative of a function differentiable on an interval cannot have any simple discontinuities

Question

Here is Theorem 5.12 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is a real differentiable function on $[a, b]$ and suppose $f^\prime(a) < \lambda < f^\prime(b)$. Then there is a point $x \in (a, b)$ such that $f^\prime(x) = \lambda$.

A similar result holds of course if $f^\prime(a) > \lambda > f^\prime(b)$.

And, here is the Corollary by Rudin to the above theorem.

If $f$ is differentiable on $[a, b]$, then $f^\prime$ cannot have any simple discontinuities on $[a, b]$.

Finally, here is Definition 4.26 in Baby Rudin, 3rd edition:

Let $f$ be defined on $(a, b)$. If $f$ is discontinuous at a point $x$, and if $f(x+)$ and $f(x-)$ exist, then $f$ is said to have a discontinuity of the first kind, or a simple discontinuity, at $x$. Otherwise the discontinuity is said to be of the second kind.

There are two ways in which a function can have a simple discontinuity: either $f(x+) \neq f(x-)$ [in which case the value $f(x)$ is immaterial], or $f(x+) = f(x-) \neq f(x)$.

I've just found out that this very question has an answer at the following link.

Discontinuities of the derivative of a differentiable function on closed interval

However, I'm stuck on the following part of the answer.

For the the second kind, say $f'(x-)<f'(x+)$ and pick $\lambda\in(f'(x-),f'(x+))$, and let $u<x<v$ with $u$, $v$ sufficiently close to $x$ so that $f'(z)<\lambda$ for $z\in[u,x)$, and $f'(z)>\lambda$ for $z\in(x,v]$.

Why can we not have $f^\prime(x) = \lambda$ in this case (and thus fail to get our desired contradiction)?

In order to demonstrate my understanding of the answer, I'll rephrase it, or rather expand upon it.

Suppose $f$ is a real function which is differentiable on a closed interval $[a, b]$, and suppose $f^\prime$ has a simple discontinuity at a point $p \in [a, b]$.

Then as both $f^\prime(p+)$ and $f^\prime(p-)$ exist, so we must have $p \in (a, b)$. Am I right?

Now there are two possible cases, according to whether (i) $f^\prime(p-) = f^\prime(p+) \neq f^\prime(p)$ or (ii) $f^\prime(p-) \neq f^\prime(p+)$.

Case (i): We can assume without loss of generality that $f^\prime(p-) = f^\prime(p+) < f^\prime(p)$. Let $\lambda$ be a real number such that $$\lambda \in \left( \ \lim_{x \to p} f^\prime(x), \ f^\prime(p) \ \right);$$ that is, $$ \lim_{x \to p} f^\prime(x) < \lambda < f^\prime(p).$$ Now let $\varepsilon$ be any real number such that $$0 < \varepsilon < \lambda - \lim_{x \to p} f^\prime(x).$$ Then we can find a real number $\delta > 0$ such that $$ (p-\delta, p+\delta) \subset (a, b), $$ and $$ \left\vert f^\prime(x) - \lim_{x \to p} f^\prime(x) \right\vert < \varepsilon$$ for all $x \in (p-\delta, p+\delta) \setminus \left\{ p \right\}$; that is, $$ \lim_{x \to p} f^\prime(x) - \varepsilon < f^\prime(x) < \lim_{x \to p} f^\prime(x) + \varepsilon $$ for all $x \in (p-\delta, p+\delta) \setminus \left\{ p \right\}$.

But $$ \lim_{x \to p} f^\prime(x) + \varepsilon < \lambda.$$ So we can conclude that $$ f^\prime(x) < \lambda \ \mbox{ for all } \ x \in (p-\delta, p+\delta) \setminus \left\{ p \right\}.$$ Let us put $$y \colon= p - \frac{\delta}{2}, \ \mbox{ and } \ z \colon= p + \frac{\delta}{2}.$$ Then we can conclude that $$ f^\prime(x) < \lambda \ \mbox{ for all } \ x \in [y, p). \tag{1} $$ Thus in particular, $$f^\prime(y) < \lambda.$$
But we have assumed that $$\lambda < f^\prime(p). $$ So we must have a point $x \in (y, p)$ for which $f^\prime(x) = \lambda$, which contradicts (1) above. [We can of course take the argument forward with $z$ instead of $y$.] Am I right?

Case (ii): We can assume without loss of generality that $f^\prime(p-) < f^\prime(p+)$. Let $\lambda$ be a real number such that $$ f^\prime(p-) < \lambda < f^\prime(p+).$$ Let's choose a real number $\varepsilon$ such that $$0 < \varepsilon < \min \left\{ \ f^\prime(p+) - \lambda, \ \lambda - f^\prime(p-) \ \right\}. \tag{2} $$ For this $\varepsilon$, we can find positive real numbers $\delta_1$ and $\delta_2$ such that $$(p, p+\delta_1) \subset (a, b), \ \mbox{ and } \ (p-\delta_2, p) \subset (a, b),$$ and also $$ \left\vert f^\prime(x) - f^\prime(p+) \right\vert < \varepsilon \ \mbox{ whenever } \ x \in (p, p+\delta_1),$$ and $$ \left\vert f^\prime(x) - f^\prime(p-) \right\vert < \varepsilon \ \mbox{ whenever } \ x \in (p-\delta_2, p).$$ That is, $$ f^\prime(p+) - \varepsilon < f^\prime(x) < f^\prime(p+) + \varepsilon \ \mbox{ whenever } \ x \in (p, p+\delta_1), \tag{3} $$ and $$ f^\prime(p-) - \varepsilon < f^\prime(x) < f^\prime(p-) + \varepsilon \ \mbox{ whenever } \ x \in (p-\delta_2, p). \tag{4}$$ But from (2), we note that $$f^\prime(p-) + \varepsilon < \lambda < f^\prime(p+) - \varepsilon. \tag{5} $$ From (3) and (5) we conclude that $$ \lambda < f^\prime(x) \ \mbox{ whenever } \ x \in (p, p+\delta_1).$$ And, from (4) and (5) we conclude that $$f^\prime(x) < \lambda \ \mbox{ whenever} \ x \in (p - \delta_2, p). $$ Then if we put $$y \colon= p-\frac{\delta_2}{2} \ \mbox{ and } \ z \colon= p + \frac{\delta_1}{2},$$ then we conclude that $$ \lambda < f^\prime(x) \ \mbox{ whenever } \ x \in (p, z ], \tag{6} $$ and $$f^\prime(x) < \lambda \ \mbox{ whenever} \ x \in [y, p). \tag{7}$$ In particular, $$f^\prime(y) < \lambda < f^\prime(z).$$ So there is some point $x \in (y, z)$ such that $$f^\prime(x) = \lambda, $$ which in the light of (6) and (7) implies that $$f^\prime(p) = \lambda.$$

Is my argument up to this point correct? If so, then where do we get our desired contradiction?

Answer 1

I think you're right, the proof by Harald has a small gap.

It is easy to fix. Suppose $f'(x-) < f'(x+)$. Choose $\lambda_1, \lambda_2$ with $f'(x-) < \lambda_1 < \lambda_2 < f'(x+)$. Then we can find some $\lambda \in (\lambda_1, \lambda_2)$ such that $\lambda \ne f'(x)$.

Now there exist $u < x < v$ such that $f'(z) < \lambda_1$ on $[u,x)$ and $f'(z) > \lambda_2$ on $(x, v]$. In particular, taking any $a \in [u,x)$ and any $b \in (x,v]$, we have $f(a) < \lambda_1 < \lambda < \lambda_2 < f(b)$, yet by construction there is no $z \in [a,b]$ with $f(z) = \lambda$.