# Show that for any $x_0\in \mathbb{R}$, the one sided limits exist and that $f^+(x_0)\geq f^-(x_0)$.

Suppose $f(x)$ is a monotone increasing function defined for all $x\in \mathbb{R}$. Show that for any $x_0\in \mathbb{R}$, the one sided limits $$f^+(x_0)=\lim_{x\to x_0^+}f(x) \text{ and } f^-(x_0)=\lim_{x\to x_0^-}f(x)$$ exist and that $f^+(x_0)\geq f^-(x_0)$.

I know that the limit definition is: $\forall \epsilon>0$ $\exists \delta>0$ such that $|f(x)-f^+(x_0)|<\epsilon$ where $|x-x_0|<\delta$.

I know that $x_0<x$, so $f(x_0)<f(x) \Rightarrow f(x)-f(x_0)>0$ $$|f(x)-f^+(x_0)|=|f(x)-f(x_0)+f(x_0)-f^+(x_0)|\leq |f(x)-f(x_0)|+|f(x_0)-f^+(x_0)|$$

But where do I go from here?

• You switched the quantifiers for the limit definition. – Fundamental Feb 9 '16 at 3:08
• $$f(x) > f(x_{0}) \implies f(x) - f(x_{0}) > 0$$ – mattos Feb 9 '16 at 3:10
• @UsernameUnknown You've changed it, but now you have changed the other statement $f(x) > f(x_{0})$ to $f(x_{0}) > f(x)$, which now makes the statement incorrect again. It should read $$f(x) > f(x_{0}) \implies f(x) - f(x_{0}) > 0$$ – mattos Feb 9 '16 at 3:15
• @UsernameUnknown Now it is fine. – mattos Feb 9 '16 at 3:18

If $y < x_{0} < z$ then we have (by the assumptions given in question) $$f(y) \leq f(x_{0}) \leq f(z)\tag{1}$$ Now consider the set $$A = \{f(x) \mid x > x_{0}\}\tag{2}$$ then clearly $A$ is bounded below by $f(x_{0})$ and hence $\inf A = a$ exists and $a \geq f(x_{0})$. Similarly if $B = \{f(x) \mid x < x_{0}\}$ then $b = \sup B$ exists and $b \leq f(x_{0})$. Thus we have $$b = \sup B \leq f(x_{0}) \leq \inf A = a\tag{3}$$
We show that $$f^{+}(x_{0}) = \lim_{x \to x_{0}^{+}}f(x) = a = \inf A\tag{4}$$ To show this let us take any arbitrary $\epsilon > 0$. Since $a = \inf A$, it follows by definition of infimum that there is a member $f(x') \in A$ such that $f(x') < a + \epsilon$. This $x' > x_{0}$ by definition and let's set $\delta = x' - x_{0} > 0$. Now we can see that if $x$ is any number with $0 < x - x_{0} < \delta$ then $x_{0} < x < x'$ and hence $f(x_{0})\leq f(x)\leq f(x') < a + \epsilon$ and clearly by definition of $A$ and $\inf A = a$ we have $a \leq f(x)$. It follows that we have $|f(x) - a| < \epsilon$ for all $x$ with $0 < x - x_{0} < \delta$. It is now clear that equation $(4)$ is established. In the same manner we can establish that $$f^{-}(x_{0}) = \lim_{x \to x_{0}^{-}}f(x) = b = \sup B\tag{5}$$ and from $(3)$ we see that $b \leq a$ so that $f^{-}(x_{0}) \leq f^{+}(x_{0})$.
Let $f^+(x_0)=\inf_{x \in (x_0, \infty)} f(x)$, this is our candidate limit point. By the definition of $\inf$, we can find an for any $\epsilon>0$, there is an $x$ s.t. $f(x)<f^+(x_0) + \epsilon$, and since $f(x)$ is monotonically increasing, this holds for all $x \in (x_0,x)$. Similar logic shows this in the other direction.
If the limit from below was above the limit from above, then we could find a sufficiently small $\epsilon$ so that there exist $x<x_0$ and $y>x_0$ such that $|f^-(x_0)-f(x)|<\frac{\epsilon}{2}$, $|f^+(x_0)-f(y)|<\frac{\epsilon}{2}$, and $|f^+(x_0)-f^-(x_0)|>\epsilon$. This would then mean that $f(x)>f(y)$, which contradicts the monotonicity assumption.
• Real numbers satisfy the least upper bound property (and greatest lower bound), so the inf over a set that's bounded below exists (in this case $\inf_{x \in (x_0, \infty)} f(x)$). We then confirmed that this was the limit, by showing that for any $\epsilon$, we could find an $x$ s.t. $|f(x)-f^+(x_0)|<\epsilon$, so setting $\delta <x-x_0$ gives us our $\delta$. – George Feb 9 '16 at 3:35