Prove $f$ is decreasing?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function on $[a,+\infty)$, such that $f^\prime$ is increasing and $\lim_{x\to\infty}f(x)=0$ I have two questions as follows

(1) how do you prove that $f$ is decreasing on $[a,+\infty)$

(2) Can the condition "the sequence $f(n)$ convergent to zero" work instead of $\lim_{x\to\infty}f(x)=0$?

I tried to prove this by definition of derivative as follows

Since $f^\prime(a)\leq f^\prime(x)$ when $a\leq x_0\;;x_0\in[a,+\infty)$ then we have $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\leq\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$$ but this way can not work.

Answer 1

With a little work, you can prove that $\lim_{x\to\infty}f(x)=0$ implies that $\lim_{x\to\infty}f'(x)=0$. Intuitively this should be clear, since $\lim_{x\to\infty}f(x)=0$ means the graph has a horizontal asymptote, so its derivative eventually tapers off to $0$ as the graph becomes flat.

Now, since $f'(x)$ is increasing and its limit is $0$, it must be negative. Which is exactly what you want to prove for $(1)$.

For $(2)$, the answer is "yes". Just write out the definitions of convergence and what it means for $\lim_{x\to\infty}f(x)=0$ and they should match up enough to see this.

Answer 2

(1) Assume $a\le u<v$. Let $s=\frac{f(v)-f(u)}{v-u}$. By the MVT, there exists $\xi \in (u,v)$ with $f'(\xi)=s$. Likewise, for $x>v$, thee exists $\eta\in(v,x)$ such that $\frac{f(x)-f(v)}{x-v}=f'(\eta)$. As $f'$ is decreasing, we conclude $f'(\eta)\le s$ and so $$f(x)\le s(x-v)+f(v) \qquad \forall x>v$$ If $s<0$ then for arbitrary $M\in\Bbb R$, this implies $$\tag{$\star$} f(x)\le -M\qquad\forall x>\max\{v,\tfrac{M-f(v)}s+v\}$$ and this contradicts $\lim_{x\to\infty}f(x)=0$ if we consider any positive $M$. Therefore $s\ge0$

(2) Our result $(\star)$ leads to a contradiction under much weaker conditions than $\lim_{x\to\infty}f(x)=0$, for example $$\liminf_{x\to\infty} f(x)<\infty$$ And of course $\lim_{n\to\infty}f(n)=0$ implies $\liminf_{x\to\infty} f(x)<0$.

Answer 3

Let me re-phrase the conditions: Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a continuous function such that

1. $f'$ exists on $(a,\infty)$ and is monotone increasing (non-strict sense).

2. $\lim_{n\rightarrow\infty} f(n) =0$.

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Claim 1: For each $x\in(a,\infty)$, $f'(x)\leq 0$.

Prove by contradiction. Suppose that there exists $x_0\in(a,\infty)$ such that $f'(x_0)>0$.

By condition 2, for $\varepsilon = f'(x_0)/4>0$, there exists $N$ such that $|f(n)|<\varepsilon$ whenever $n\geq N$. Choose an integer $n$ such that $n> \max(N, x_0)$. On one hand, $|f(n+1)-f(n)|<2\varepsilon = \frac{1}{2}f'(x_0)$. On the other hand, by the mean value theorem, $|f(n+1)-f(n)|=|f'(\xi) [(n+1)-n]| \geq f'(x_0)$ because $\xi\in(n,n+1)\Rightarrow 0<f'(x_0)\leq f'(\xi)$ by Condition 1. Obviously, this is a contradiction.

By Claim 1, $f'\leq 0$ which implies that $f$ is decreasing on $[a,\infty)$.

Answer 4

Since $f'$ is increasing therefore either $f'(x) $ tends to a limit $L$ or to $\infty$ as $x\to\infty $. By L'Hospital's Rule the ratio $f(x) /x$ also tends to either $L$ or $\infty$. Since we are given that $f(x) \to 0$ it follows that $L=0$ and thus $f'(x) \to 0$ as $x\to\infty $. Since $f'$ is increasing its values do not exceed the limit $0$ and therefore $f'(x) \leq 0$ for all $x\in[a, \infty) $ and thus $f$ is decreasing in $[a, \infty) $.

You can see that the above argument works even if we are only given that sequence $f(n) \to 0$. The same proof and conclusion applies if we are given that $f(x) $ or $f(n) $ tends to a constant value $k$.

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You should observe that the definition of derivative is local ie it deals with the behavior of a function in the neighbourhood of a single point and hence we can't obtain any global information (like behavior on an interval) about a function using just the definition of derivatives. To make such inferences one needs to use theorems which deal with global behavior eg mean value theorem (or theorems related to completeness of real numbers). My approach uses L'Hospital's Rule which is itself based on Cauchy's Mean Value Theorem.

Also note that L'Hospital's Rule is a powerful tool for theoretical investigation (as used in the proof above). It is rather tragic that for most beginners the rule is more like "differentiate and plug" mantra for evaluation of limits.

Answer 5

$f'$ is increasing implies $f''>0$, which in turn implies the function is convex.

$\lim_\limits{x\to\infty}f(x)=0$ implies the function is decreasing, because $y=0$ is the horizontal asymptote.