If $a<b$, the $\lim_{x\to a^+}f(x) \le \lim_{x\to b^-}f(x) $?

Question

Let $f$ be a monotone non-decreasing function real valued function on $\mathbb R$.

To Prove:If $a<b$, the $\lim_{x\to a^+}f(x) \le \lim_{x\to b^-}f(x) $

PROOF:If $0<x<1$ then $f(x)=bx+(1-x)a$ which is a non decreasing function.

Now,$\lim_{x\to a^+}f(x) - \lim_{x\to b^-}(x) =ab+(1-a)a-b^2-(1-b)a=a^2-b^2\le0$ .Hence,$\lim_{x\to a^+}f(x) \le \lim_{x\to b^-}(x) $

I'm not getting how to prove it for $(0,1)^c$.

Answer 1

• You cannot prove true a general statement by just trying out an example.
• You can prove false a general statement by showing an example where the statement doesn't hold.

The statement you need to prove has several parts. First, you have to show that both $$ \lim_{x\to a^+}f(x) \qquad\text{and}\qquad \lim_{x\to b^-}f(x) $$ exist and then that the former is less than or equal to the latter.

For the existence, the hint is to prove that $$ \lim_{x\to a^+}f(x)=\inf\{f(x):x>a\} $$ and similarly for the other one.

Let $l=\inf\{f(x):x>a\}$ and take $\varepsilon>0$. Then there exists $x_0>a$ with $f(x_0)<l+\varepsilon$. Let $\delta=x_0-a$. Since $f$ is nondecreasing, for $a<x<x_0=a+\delta$, it holds $0\le f(x)-l<\varepsilon$.

Next you want to show that $$ \inf\{f(x):x>a\}\le\sup\{f(x):x<b\} $$ Hint: $f(a)\le f(m)\le f(b)$, where $m=(a+b)/2$.

Answer 2

One way is contradiction. Assume the limits as $A, B$ and let $A>B$. Now we can choose an $x$ near $a$ and $x'$ near $b$ such that $a<x<x'<b$ and $f(x) $ is near $A$ and $f(x') $ is near $B$. Since $A>B$, it is possible to choose $x, x'$ such that $f(x) >f(x') $ and this contradicts that $f$ is non-decreasing.

If you prefer direct approach then choose $c$ such that $a<c<b$ and if $x, t$ are such that $a<x<c<t<b$ then $f(x) \leq f(c) \leq f(t) $. Now letting $x\to a^{+} $ and $t\to b^{-} $ we get the desired inequality. Note that this uses the result that limits preserve inequalities but weaken them.