let $f: X\to\mathbb R$ monotonous. If $(x_n)$ exists in $X$ with $x_n>a$, $\lim x_n = a$ and $\lim f (x_n)=L$, then $\lim f(x)=L$ when $x\to a^+$

Question

Let $f: X \to \mathbb{R}$ be monotonic and $a \in X^\prime_+$. Then if there is a sequence of points $x_n \in X$ with $x_n > a$, $\lim x_n = a$ and $\lim f(x_n) = L$, we have $\lim_{x \to a^+} f(x) = L$.

Answer 1

Suppose, on the contrary, that $f(x)$ does not converge to $L$ from the right. Then we can construct a sequence $\{ y_n \}$ such that $y_n > a, y_n \to a$ but $\lvert f(y_n) - L \rvert \geq \epsilon$ for some value $\epsilon > 0$. In other words, we pick the terms $y_n$ that, no matter how close to $a$, still can't get $f(y_n)$ arbitrarily close to $L$.

Since $f(x_n) \to L$, for large enough $n$, fixing some $N$, we have $\lvert f(x_N) - L \rvert < \epsilon$. Since $y_m$ converges to $a$, we can pick an index $M$ big enough so that $y_M$ is closer to $a$ than $x_N$. Now, since $x_n \to a$ too, we pick a third index $K$ large enough so that $\lvert f(x_K) - L \rvert < \epsilon$ and $x_K$ is closer to $a$ than $y_M$.

This way we get

$$a < x_K < y_M < x_N \; .$$

Now, since $f$ is monotonic, either $f(x_K) \leq f(y_M) \leq f(x_N)$ or the opposite. Suppose the first case is true, for in the second we could use the same reasoning. Then remeber $\lvert f(x_N) - L \rvert < \epsilon$ also means

$$L - \epsilon < f(x_N) < L + \epsilon \; ,$$

and the same is true for $x_K$.

Hence we have

$$ L - \epsilon < f(x_K) \leq f(y_M) \leq f(x_N) < L + \epsilon \; ,$$ which is equivalent to $\lvert f(y_M) - L \rvert < \epsilon$, but this is a contradiction since we constructed $\{ y_n \}$ with the property that $\lvert f(y_n) - L \rvert \geq \epsilon$ .

Answer 2

Let $f: X \to \mathbb{R}$ be a monotonic function and $a \in X^\prime_+$.

If there is a sequence of points $x_n \in X$ with $x_n > a$, $\lim_\limits{n\to\infty}x_n = a$ and $\lim_\limits{n\to\infty} f(x_n) = L$, then we have that $\lim_\limits{x \to a^+} f(x) = L$.

We are going to prove this property in the case that $f$ is a non-decreasing function and $L$ is a real number.

Let $\epsilon$ be any positive real number.

Since $\lim_\limits{n\to\infty} f(x_n) = L$, there exists $\nu_1\in\mathbb{N}$ such that $L-\epsilon<f(x_n)<L+\epsilon\;\;$ for all $\;\;n\ge\nu_1.\;\;\;\;\;(*)$

Let $\delta=x_{\nu_1}-a>0$.

We have to prove that $\;\forall x\in X\cap\left]a,a+\delta\right[\;$ results $\;L-\epsilon<f(x)<L+\epsilon$.

Let $x$ be any real number in $X\cap\left]a,a+\delta\right[.$

It implies that $\;x<a+\delta=x_{\nu_1}$.

Since $\lim_\limits{n\to\infty}x_n = a$, there exists $\nu_2\in\mathbb{N}$ such that $\;\;a<x_n<x\;\;$ for all $n\ge\nu_2$.

Let $\nu_3=\max\{\nu_1,\nu_2\}$, it results that $x_{\nu_3}<x<x_{\nu_1}$.

From $(*)$ and by using the fact that $f$ is a non-decreasing function, we get

$L-\epsilon<f\left(x_{\nu_3}\right)\le f(x)\le f\left(x_{\nu_1}\right)<L+\epsilon$.

So we have proved that

$\forall\epsilon>0$ there exists $\delta>0$ such that $\;\forall x\in X\cap\left]a,a+\delta\right[\;$ results $\;L-\epsilon<f(x)<L+\epsilon$

that is

$\lim_\limits{x \to a^+} f(x) = L$.

By proceeding analogously, it is possible to prove this property in the case that $f$ is decreasing or increasing or non-increasing and $L$ is $+\infty$ or $-\infty$.