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Let $f\in C[a,b]$, and except $\{x_n\}\subset [a,b]$, $f'(x)>0$. Show that $f$ is strictly increasing.

Note that we do not know the differentiability of $f$ at each $x_n$. What we know is just: $f'(x)>0, \forall\ x\not\in \{x_n\}$.

If ``except $\{x_1,\cdots,x_n\}\subset [a,b]$, $f'(x)>0$, then it is OK, but dividing the interve into finite number of small intevels, on each part $f$ is strictly increasing.

The trouble is $\{x_n\}$, a sequence...Any ideas?

xldd
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  • Probably something to do with the measure of the sequence. Since the singletons ${x_i}$ are of zero measure, then their union is also of null measure, and so, $f'$ is stricly positive almost everywhere except on a set of null measure. Then $f$ is strictly increasing almost everywhere. Just a thought. – Hamdiken Jul 16 '22 at 00:53
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    @Hamdiken This argument doesn't hold water I'm afraid. Consider the Cantor function (denote it by $C$). Then the function $f$ defined by $f(x) = x - 2 C(x)$ is differentiable almost everywhere with $f' > 0$ almost everywhere, but f(0) > f(1). – Sam Forster Jul 16 '22 at 01:04
  • @SamForster You're right. Your comment serves as a counterexample to OP's question. – Hamdiken Jul 16 '22 at 01:12
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    @Hamdiken It's not a counter-example actually, because the Cantor set is uncountable. OP's question is actually true, if there are only countably many exceptions, then $f$ is strictly increasing. – Sam Forster Jul 16 '22 at 01:14
  • @SamForster So what I said was actually true, except that I had to specify that the null measure set had to be countable ? – Hamdiken Jul 16 '22 at 01:18
  • @Hamdiken Exactly. But since every countable set is a null set, saying countable is enough. Saying "a null set which is countable" is kind of redundant. – Sam Forster Jul 16 '22 at 01:22
  • @SamForster I see now. Thanks for the explanation ! – Hamdiken Jul 16 '22 at 01:23

2 Answers2

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We first show that $f(a)\leq f(b)$. Suppose that $f(a) > f(b)$ and select $c$ such that $f(a)>c>f(b)$ and $f(x_n)\neq c$ for all $n\in \mathbb{N}$. Consider the set $$A=\{x\in [a,b] : f(x) > c\}$$ and let $\beta = \sup A$. Now if we take $y_n \in A$ with $y_n \to \beta$ and pass to the limit we get $f(\beta)\geq c$. But if $f(\beta)>c$ for sufficiently small $\epsilon > 0$ we would have $f(\beta + \epsilon)>c$, a contradiction. So we have established that $f(\beta)=c$. Now look at $$f'(\beta) = \lim_{x\to \beta}\frac{f(x)-f(\beta)}{x-\beta}$$ For $x>\beta$ we have $f(x)\leq c$ and $f(\beta)=c$ so it must be the case that $f'(\beta) \leq 0$ (we know that $f'(\beta)$ exists because we required that $f(x_n)\neq c$ for all $n\in \mathbb{N}$). This is a contradiction since we have $f'>0$ everywhere except possibly the $x_n$'s. So $f(a)\leq f(b)$. Now for all $x\in (a,b)$ we have $f(a)\leq f(x) \leq f(b)$ by repeating the above argument. If $f(a)=f(b)$ then $f$ would be constant and $f'=0$ so $f(a)<f(b)$.

Now for arbitrary $x,y\in [a,b]$ with $x<y$ we can repeat our argument in $[x,y]$ to get $f(x)<f(y)$, thus proving that $f$ is strictly increasing.

George Giapitzakis
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Comment only.

The idea in the proof that Giorgos has given is worth pondering over. If there were no points at which $f'(x)$ failed to be positive, then (using the intermediate value property of continuous functions) select any $c\in (a,b)$ with $f(a)>c>f(b)$ and use his "last point argument" to consider $$ \beta =\sup\{ x\in (a,b): f(x)>c\}$$ where we are sure that $f'(\beta)>0$ etc.

If there is a set of points $E\subset (a,b)$ where we perhaps do not have a positive derivative, well... just avoid them. If $E$ is countable, so is $f(E)$ and there is plenty of room to choose a value $c\not\in f(E)$. That will get us a $\beta\not\in E$ etc.

That method leads to an interesting and simple generalization using the same method, due to Antoni Zygumund. As a follow up to this "exercise" I recommend proving this theorem.

Theorem (Zygmund) Let $f$ be a continuous function such that the set of values assumed by $f(x)$ at the points $x$ where the upper right Dini derivative $D^+f(x)\leq 0$ contains no nondegenerate interval. Then $f$ is monotone nondecreasing.

B. S. Thomson
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