Positive derivative implies increasing without Mean Value Theorem

Question

The result below is usually proven by using the Mean Value Theorem (see e.g. ProofWiki).

But can we also prove it more directly (and fairly elegantly) without resort to the MVT?

Suppose $f:[a,b]\rightarrow \mathbb{R}$ is differentiable.

• If $f'(x)\geq0$ for all $x\in[a,b]$, then $f$ is increasing on $[a,b]$.
• If $f'(x)>0$ for all $x\in[a,b]$, then $f$ is strictly increasing on $[a,b]$.
• If $f'(x)\leq0$ for all $x\in[a,b]$, then $f$ is decreasing on $[a,b]$.
• If $f'(x)<0$ for all $x\in[a,b]$, then $f$ is strictly decreasing on $[a,b]$.

Answer 1

This old one popped up today. So, if there is some interest here is a contribution. As @user333083 points out, if you really want to bypass the MVT, just come up with a simple compactness argument (as did @Paul). My favorite is Cousin's lemma [1], which I will review for those who haven't seen it. [It dates to the 1890s.]

Let $\cal{C}$ be a collection of closed subintervals of $[a,b]$. We say

1. $\cal C$ is additive if, whenever $[u,v]$ and $[v,w]$ belong to $\cal C$, then so to does the interval $[u,w]$.

2. $\cal C$ is a full cover of $[a,b]$ if, for every $x\in [a,b] $ there exists a $\delta>0$ so that every interval $[c,d]\subset [a,b]$ with the property that $x\in [c,d]$ and $d-c<\delta$ must belong to $\cal C$.

Cousin covering lemma: Suppose that $\cal C$ is an additive, full cover of an interval $[a,b]$. Then $\cal C$ contains every closed subinterval of $[a,b]$.

Proof. Follows easiest from the nested interval theorem, although you can base it on Heine-Borel or Bolzano-Weirstrass. If there is an interval $[a_1,b_1]$ not in $\cal{C}$ then split it in half and choose the half that also is not in $\cal{C}$, call it $[a_2,b_2]$, etc.

----------Proof of monotonicity theorem--------------

Definition. $f$ is locally strictly increasing at a point $x\in [a,b]$ if there is a $\delta>0$ so that $$f(y_1)<f(x) < f(y_2)$$ for all points $y_1$ and $y_2$ in the interval that satisfy $x-\delta< y_1<x<y_2< x +\delta$.

Note that if $f'(x)>0$ at a point $x$ then certainly $f$ is increasing at that point. [Warning: this is very different from asserting $f$ is actually increasing on the whole interval $(x-\delta,x+\delta)$.]

Theorem. If $f:[a,b]\to \mathbb R$ is locally strictly increasing at each point of $[a,b]$ then $f$ is strictly increasing on $[a,b]$.

Proof. Let $\cal{C}$ be the collection of all subintervals $[c,d]\subset [a,b]$ for which $f(c)<f(d)$. This is clearly additive and, because of the local assumption, it is a full cover. Thus $\cal{C}$ contains all subintervals of $[a,b]$. QED

Corollary. If $f:[a,b]\to \mathbb R$ satisfies $f'(x)>0$ at each point of $[a,b]$ then $f$ is strictly increasing on $[a,b]$.

Proof. Follows directly from the theorem.

REFERENCE.

[1] Cousin, P. (1895). Sur les fonctions de $n$ variables complexes. Acta Mathematica. 19(1):1--61.

Answer 2

Long comment: If you want to learn to prove things about differentiable functions then this is the wrong question! Because MVT is the way things about differentiable functions are proved.

Of course there must exist counterexamples to that last statement, but regardless you'll be better off if, when you need to prove something about diifferentiable functions, you automatically consider whether you can apply MVT.

If you want to master elementary-calculus-with-proofs you should

1. Study the proof of Rolle's theorem until it seems "obvious".

2. Similarly for the proof that Rolle implies MVT.

3. Similarly for the proof of the current result using MVT.

Honest. If, as seems possible, I'm better at proving things about derivatives than you are, the reason is I did (1), (2) and (3) long ago. Wondering how you can avoid MVT here is not going to be nearly as useful.

Answer 3

Absolutely yes!!

I've always thought that using MVT in these proof is kinda like cheating, the reason is that I don't find these proofs to be intuitive, and also in some cases they're hard to generalize. I state that this alternative proof might seem so difficult as to lead us to ask ourselves why don't just stick to the classic MVT's proof. In my opinion the reason is that this proof is based on an idea that can be repeated in other proofs in which the MVT may not be used, more generally I find it very instructive to prove propositions in very different ways.

Now let's dive into the proof

Let $f \; : \; ]a,b[ \; \to \mathbb{R}$ be a everywhere differentiable function with $$f'(x) > 0 \hspace{0.3cm} \forall x \in ]a,b[$$

then $f(\alpha) < f(\beta)$ for all $\alpha < \beta$ in $]a,b[$

(I'm taking an open interval just for simplicity but it's not hard to generalize the proof for a closed interval)

the idea of the proof is, let $\alpha < \beta$, then

$$f(\beta) - f(\alpha) = \int_\alpha^\beta{f'(t)dt} > 0$$

The problem with this is that $f'(t)$ could be non Riemann-integrable. To fix this we need to replace the Riemann-integral with something "similar" but that always exists (for the function $f$). You can either use the Henstock-Kurzweil integral but it would be too advanced, so I thought of replacing the Riemann-Integral with a Riemann Sum, the idea is to do something like this

$$f(\beta) - f(\alpha) = \sum_{k=1}^{n}{(f(x_k) - f(x_{k-1}))}$$ where

$f(x_k) - f(x_{k-1})$ is always positive (or bigger than a certain $-\epsilon$) because $f(x_k) - f(x_{k-1}) = \frac{ f(x_k) - f(x_{k-1}) }{x_k - x_{k-1}}(x_k - x_{k-1})$ and $\frac{ f(x_k) - f(x_{k-1}) }{x_k - x_{k-1}} \approx f'(x_k) > 0$ because $x_k - x_{k-1}$ is "small"

formalizing this proved to be more difficult than expected, at the end I managed to do it using the compactness of $[\alpha,\beta]$.

This is how I did it

Proof

Let $\alpha < \beta$ be in $]a,b[$, I want to show that $f(\beta) - f(\alpha) > 0$

let $x \in [\alpha,\beta]$, i know that $$\lim_{y \to x}{\frac{f(x)-f(y)}{x - y}} > 0$$

therefore there exists $\delta_x > 0$ such that

$$\frac{f(x) - f(y)}{x - y} > 0 \;\; \text{ whenever } \;\; y \in B_{\delta_x}(x)$$

it's pretty clear that $\{ B_{\delta_x}(x) \; : \; x \in [\alpha,\beta] \}$ is an open cover of the compact set $[\alpha,\beta]$, therefore it admits a finite subcover, meaning that there exists $\mathcal{F} \subset [\alpha,\beta]$ such that $\mathcal{F}$ is finite and

$$[\alpha,\beta] \subset \cup_{x \in \mathcal{F}}{B_{\delta_{x}}(x)}$$

whenever $y \in [\alpha,\beta]$ let $\mathcal{F}(y) := \{ x \in \mathcal{F} \; : \; y \in B_{\delta_x}(x) \}$

let $x_1$ be such that $x_1 \in \mathcal{F}(\alpha)$ and $\delta_{x_1} = \max\{ \delta_x \; : \; x \in \mathcal{F}(\alpha)\}$

if $x_1 + \delta_{x_1} > \beta$ I'm done, otherwise

let $x_3$ be such that $x_3 \in \mathcal{F}(x_1 + \delta_{x_1})$ and $\delta_{x_3} = \max\{ \delta_x \; : \; x \in \mathcal{F}(x_1 + \delta_{x_1})\}$

I know that $x_1 < x_3$ because if $x_3 \geq x_1$ then $x_1 + \delta_{x_1} \in B_{\delta_{x_3}}(x_3) \implies \delta_{x_3} > \delta_{x_1} \text{ and } \alpha \in B_{\delta_{x_3}}(x_3)$ which is a contradiction by the definition of $x_1$.

so i know that $]x_1,x_3[ \cap B_{\delta_{x_1}}(x_1) \cap B_{\delta_{x_3}}(x_3)$ is non empty. let $x_2 \in ]x_1,x_3[ \cap B_{\delta_{x_1}}(x_1) \cap B_{\delta_{x_3}}(x_3)$

If $x_3 + \delta_{x_3} > \beta$ I'm done, otherwhise I keep doing this, in general once i have defined

$x_1,x_2,\dots,x_{2n - 1}$ if $x_{2n - 1} + \delta_{x_{2n-1}} > \beta$ I'm done, otherwise I let

$x_{2n + 1}$ be such that $x_{2n + 1} \in \mathcal{F}(x_{2n - 1} + \delta_{x_{2n-1}})$ and $\delta_{x_{2n+1}} = \max\{ \delta_x \; : \; x \in \mathcal{F}(x_{2n - 1} + \delta_{x_{2n-1}}) \}$

and I let $x_{2n} \in ]x_{2n - 1},x_{2n+1}[ \cap B_{\delta_{x_{2n-1}}}(x_{2n-1}) \cap B_{\delta_{x_{2n+1}}}(x_{2n+1})$

by keeping doing this at a certain point I must end (otherwise the sequence $x_1 < x_3 < \dots < x_{2n-1},\dots$ will be an infinite subset of $\mathcal{F}$, which is a finite set), therefore I find

$x_1 < x_2 < \dots < x_{2n} < x_{2n-1}$ be such that

$x_{2k} \in B_{\delta_{x_{2k-1}}}(x_{2k-1}) \cap B_{\delta_{x_{2k+1}}}(x_{2k+1})$

for all $k=1,2,\dots,n-1$

This means that

$$\frac{f(x_{2k})-f(x_{2k-1})}{x_{2k} - x_{2k-1}} > 0$$ $$\frac{f(x_{2k+1})-f(x_{2k})}{x_{2k+1} - x_{2k}} > 0$$

whenever $k=1,\dots,n-1$, which means that

$$\frac{f(x_{k+1}) - f(x_k)}{x_{k+1} - x_k} > 0$$

whenever $k=1,2,\dots,2n-2$

which means that

$$f(x_{k+1})-f(x_k) > 0$$

whenever $k=1,\dots,2n-1$

now I let $x_0 = \alpha$ and $x_{2n} = \beta$, I know that

$x_0 \in B_{\delta_{x_1}}(x_1)$ and $x_{2n} \in B_{\delta_{x_{2n-1}}}(x_{2n-1})$

$$f(x_{k+1})-f(x_k) > 0$$

forall $k=0,1,\dots,2n-1,2n$

so it's easy to see that

$$ f(\beta) - f(\alpha) = f(x_{2n}) - f(x_0) = \sum_{k=1}^{2n}{(f(x_k) - f(x_{k-1}))} > \sum_{k=1}^{2n}{0} = 0$$

So i have proved that

$f(\beta) - f(\alpha) > 0$

This proves the case where $f'(x) > 0$, the case where $f'(x) < 0$ can be proved in a similar way or by reasoning in terms of $-f(x)$

Now let's consider the case where $f'(x) \geq 0$

Let $\alpha < \beta$ be in $]a,b[$, let $\epsilon > 0$, let $g(x) = f(x) + \epsilon x$, then $g'(x) = f'(x) + \epsilon$ so $g'(x) > 0$ whenever $x \in ]a,b[$ therefore thanks to the previous case

$$g(\beta) - g(\alpha) > 0$$

meaning that

$$f(\beta) - f(\alpha) + \epsilon(\beta - \alpha) > 0$$

by letting $\epsilon \to 0$ I get

$f(\beta) - f(\alpha) \geq 0$

This proves the case $f'(x) \geq 0$, the case where $f'(x) \leq 0$ can be proved in a similar way or by reasoning in terms of $-f(x)$

Answer 4

Suppose by contradiction that $f'>0$ and $x<y$ and $f(x)\ge f(y).$

There exists $r>0$ such that $\forall z\in (x,x+r)\, \left(|\frac {f(z)-f(x)}{z-x}-f'(x)|<\frac {f'(x)}{2}\right).$ So there exists $r>0$ such that $\forall z\in (x,x+r)\,(f(z)>f(x)\,).$

Let $A=\{r>0:\forall z\in (x,x+r)\,(f(z)>f(x)\,)\}.$ Note that $0<r'<r\in A\implies r'\in A.$ Note that $r\in A\implies x+r\le y.$

Let $a=\sup A.$ Then $y-x\ge a>0$ and $A\supset (0,a).$ If $0<z<a$ then $z\in (0,r)$ for some $r\in A,$ so $f(x+z)>f(x).$

Claim:$ f(x+a)\le f(x)$. Proof of Claim by contradiction: If $f(x+a)>f(x)$ then there exists $\epsilon >0$ such that $\forall z\in [x+a,x+a+\epsilon)\, \left(|f(z)-f(x+a)|<\frac 1 2 (f(x+a)-f(x))\right)$

implying $\forall z\in [x+a,x+a+\epsilon)\,(f(z)>f(x)\,)$

implying $A\supset (0,a)\cup [a,a+\epsilon)=(0,a+\epsilon)$

implying $\sup A\ge a+\epsilon >a=\sup A,$ which is absurd. Therefore $ f(x+a)\le f(x)$.

Now $a>0$ so $$f'(x+a)=\lim_{z\to a;\, 0<z<a}\frac {f(x+a)-f(x+z)}{a-z} \le 0$$ because $0<z<a\implies f(x+a)\le f(x)<f(x+z).$ But we have contradicted $f'>0.$ This is the desired contradiction for proving the main result.

Answer 5

Suppose $f'(x) > 0$ for all $x \in [a, b]$. We will show that if $x < y$, then $f(x) < f(y)$. Approximation argument using $f_{\varepsilon}(x) = f(x) + \varepsilon x$ can then be used to prove the case where $f' \geq 0$.

Proof 1: Let $x \in [a, b)$ be arbitrary. By compactness, $f$ attains a minimum at some point $x_0 \in [x, b]$. We cannot have $x_0 \in (x, b)$ because if $x_0 \in (x, b]$, then since $f'(x_0) > 0$ we would have $f(x_0 - h) < f(x_0)$ for $h > 0$ small, contradicting the minimality at $x_0$. Thus $x_0 = x$. Thus $f(x) \leq f(y)$ for all $y \in [x, b]$. Since $x$ was arbitrary, this shows that $f$ is increasing. For any $x \in [a, b]$, since $f'(x) > 0$, we have $f(x + h) > f(x)$ for $h > 0$ small, so in fact $f$ is strictly increasing.

Proof 2: Let $x \in [a, b)$ be arbitrary. For contradiction, suppose that there is some $y \in (x, b]$ such that $f(y) \leq f(x)$. Let $y_0 = \inf \{y : x < y \leq b \text{ and } f(y) \leq f(x)\} \in [x, b]$. Since $f(x + h) > f(x)$ for $h > 0$ small, $y_0 > x$. By continuity, $f(y_0) \geq f(x)$. By continuity and definition of $\inf$, $f(y_0) \leq f(x)$. Hence $f(y_0) = f(x)$. But this implies that $f(y_0 - h) < f(x)$ for $h > 0$ small, so $y_0 \leq y_0 + h$, a contradiction.

Answer 6

I will first prove that if $f'(x) > 0$ for every $x \in [a, b]$ then $f$ is increasing.

To see this, suppose we define $S$ to be the set of $x \in [a, b]$ such that whenever $a \le y \le x$, we have $f(y) \ge f(a)$. Then certainly $a \in S$ so $S$ is nonempty; and also $S$ is bounded above by $b$. Therefore, we can let $c := \sup(S)$. We now claim that $c = b$. To see this, suppose to the contrary that $c < b$. Then since $f(y) \ge f(a)$ for every $y \in (a, c)$ (because $S$ must have an element greater than $y$), and $f$ is continuous, we get that $f(c) \ge f(a)$. Since $f'(c) > 0$, there exists $\delta > 0$ such that whenever $0 < |x - c| < \delta$, then $\left| \frac{f(x) - f(c)}{x - c} \right| > \frac{1}{2} f'(c) > 0$. This implies that for $c < x < c + \delta$ and $x \le b$ we have $f(x) > f(c) \ge f(a)$. But then we see that $\min(c + \frac{1}{2} \delta, b) \in S$, contradicting the fact that $c$ is an upper bound of $S$.

To conclude, we have shown that for every $x \in [a, b]$, we have $f(x) \ge f(a)$. Now, by varying $a$ in the preceding argument, we get the desired conclusion that $f$ is increasing.

(As a way to think about what this proof is doing, it uses the fact that $f'(a) > 0$ to conclude that $f(x) \ge f(a)$ for a bit above $a$. Then, at the stopping point $x_1$ of this interval, by continuity of $f$ we are still at at least $f(a)$, and using that $f'(x_1) > 0$ we can again extend a bit farther. If this ends up repeating infinitely often before reaching the right-hand side of the interval, then at the limit point of the sequence, once again by continuity we remain at $f(a)$ or above, and we can continue on once again from there. Then the supremum property of the reals implies that as long as we can continue moving onwards from any point, then we must eventually reach $b$ -- or else the supremum $c$ of all possible reachable points exists, $f(c) \ge f(a)$, we can continue onward from a bit from $c$ to get a contradiction.)

---

From here, to get to the desired statements:

(1) For each $\varepsilon > 0$, the function $g(x) = f(x) + \varepsilon x$ satisfies the hypotheses of the previous part, so $g$ is increasing. Letting $\varepsilon \to 0^+$, we get the conclusion that $f$ is increasing.

(2) By a compactness argument on $[a, b]$, $f'$ has some strictly positive lower bound $m$ on $[a, b]$. Then $g(x) = f(x) - mx$ satisfies the hypotheses of (1), so $g$ is increasing, implying that $f(x) = g(x) + mx$ is strictly increasing.

To get (3) and (4), apply (1) and (2) respectively to $g(x) = -f(x)$.