Is a function $f$ monotone decreasing if $\lim _{x\to b }f\left(x\right)\:=\:0$ and $f\left(x\right) \ge 0$ $\forall x \in [a,b]$?

Question

Let $f: [a,b] \to \mathbb{R}$.

Can we say that $f$ is monotone decreasing if $\lim _{x\to b}f\left(x\right)=0$ and $f\left(x\right) \ge 0$ $\forall x \in [a,b]$ ?

Answer 1

No, consider for example the real function $$f(x)=(x-b)^2\left(1+\sin\left(\frac{1}{x-b}\right)\right)$$ extended by continuity at $b$. Then $f(x)\geq0$ and $\lim _{x\to b }f\left(x\right)=0$, but $f$ is not decreasing in any neighbourhood of $b$ because $f(x_n)>0$ and $f(y_n)=0$ where $x_n=b+1/(2\pi n)\to b$, $y_n=b+1/(2\pi n+3\pi/2)\to 0$.

Consider $\displaystyle f(x)=\frac{1+\sin(x)}{x^2+1}$ if your $b$ is $+\infty$ or $-\infty$.

Answer 2

Assuming it was $\lim_{x\to b}f(x) = 0$ instead of $\lim_{x\to \infty}f(x) = 0$ (as $\infty \notin [a,b]$) the answer is no.

Counter example: $$ f(x) = \begin{cases} b - a- \frac{k}{n}(b-a), & \mbox{, if } x = a+ \frac{k}{n}(b-a), \forall n,k\in\mathbb{N}, \forall k\le n \\ 0, & \mbox{otherwise} \end{cases} $$

Answer 3

I suppose the limit should be as $x$ tends to $b$.

The answer to you question is NO.

You can construct a non-decreasing positive function as follows:

Consider the following points:

$x_0=a$

$x_1=a+\frac{b-a}{2}$

$x_2=a+\frac{b-a}{2}+\frac{b-a}{4}$

$x_3=a+\frac{b-a}{2}+\frac{b-a}{4}+\frac{b-a}{8}$

etc, i.e. at each step you take the midpoint of the segment on the right of you last selected point.

Set $f(x_0)=1, f(x_1)=0$. Then set $f(x_{n})=\frac{1}{2}f(x_{n-2})$. Finally, make $f$ linear in between the $x$'s.

What you now have is a piecewise linear function s.t. for a fixed $x=x_n$ for some $n$, we have that $\forall y>x, f(y)<f(x)$ though the function is not strictly decreasing.

Now easily it follows that $\lim_{x\rightarrow b}f(x)=0$.

Answer 4

No, the conditions are insufficient. This is because the condition that the limit of the function vanish as $x\to b$ is very local -- it tells us the behavior of the function only near $b$. Therefore, anything could happen as $x\to a$, for example: the function may oscillate infinitely often in a subset of $[a,b]$ and still converge to $0$ as $x\to b$.