# Show that this function is bounded

Let $$f$$ be a $$\mathbb R \rightarrow \mathbb R$$ continuous function such that : $$\lim_ {x \to \pm \infty} f(x) \in \mathbb R$$ and $$\lim_ {x \to 0} f(x) \in \mathbb R$$

How can one show that $$f$$ is bounded ? I get it "intuitively" but I cant show it rigorously

• Use the limits definition to show that it is bounded in all but a compact set, then use continuity (Weierstrass theorem in particular) to show it's bounded in all of $\mathbb{R}$ Jun 3, 2019 at 18:52
• Btw I don't think you need the existence of the limit at X=0, this follows from continuity Jun 3, 2019 at 18:55

The fact that $$\lim_{x\to0}f(x)\in\mathbb R$$ is necessarily true: since $$f$$ is continuous that limit has to be $$f(0)$$.

Now, suppose that $$f$$ is unbounded. Then, for each $$n\in\mathbb N$$, there is a $$x_n\in\mathbb R$$ such that $$\bigl\lvert f(x_n)\bigr\rvert\geqslant n$$. The sequence $$(x_n)_{n\in\mathbb N}$$ is either bounded or unbounded and:

• if it is bounded, then it has a convergent subsequence $$(x_{n_k})_{k\in\mathbb N}$$. But if $$\lim_{k\to\infty}x_{n_k}=x$$, then $$\lim_{k\to\infty}f(x_{n_k})=f(x)$$, which is impossible, since $$\bigl(f(x_{n_k})\bigr)_{k\in\mathbb N}$$ is unbounded.
• if it is unbounded, then it has a subsequence $$(x_{n_k})_{k\in\mathbb N}$$ whose limit is $$\pm\infty$$, and we then get a similar contradiction.

Juts for amusement:

Let $$\phi(x) = \begin{cases} \lim_{x \to -\infty} f(x) , & x = -{\pi \over 2} \\ f(\tan(x)), & x \in (-{\pi \over 2}, {\pi \over 2} ) \\ \lim_{x \to +\infty} f(x) , & x = {\pi \over 2} \end{cases}$$.

Show $$\phi$$ is continuous on the compact set $$[-{\pi \over 2},{\pi \over 2}]$$, hence bounded, and hence $$f$$ is bounded.

• Amusement? I think it's the superior proof. (Although "juts" is amusing)
– zhw.
Jun 3, 2019 at 19:50
• @zhw.: Just poor spelling on my part :-). Jun 3, 2019 at 19:55

If $$\lim_{x\to-\infty} f(x)=a$$ and $$\lim_{x\to\infty} f(x)=b$$ put $$|a|+|b|+1=:c$$. There is an $$M>0$$ such that $$|f(x)|\leq c$$ for all $$x\geq M$$ and all $$x\leq-M$$. Since $$f$$ is continuous there is a $$c'$$ such that $$|f(x)|\leq c'$$ for all $$x\in[-M,M]$$. It follows that $$|f(x)|\leq c+c'$$for all $$x\in{\mathbb R}$$.

The hypothesis that

$$\displaystyle \lim_{x \to \infty} f(x) \in \Bbb R \tag 1$$

means that

$$\exists L_+ \in \Bbb R, \; \displaystyle \lim_{x \to \infty} f(x) = L_+; \tag 2$$

that is,

$$\forall 0 < \epsilon_+ \in \Bbb R \; \exists 0 < M_+ \in \Bbb R, \; x > M_+ \Longrightarrow \vert f(x) - L_+ \vert < \epsilon_+; \tag 3$$

that is,

$$x > M_+ \Longrightarrow L_+ - \epsilon_+ < f(x) < L_+ + \epsilon_+; \tag 4$$

likewise the hypothesis

$$\displaystyle \lim_{x \to -\infty} f(x) \in \Bbb R \tag 5$$

gives us

$$\exists L_- \in \Bbb R, \; \displaystyle \lim_{x \to -\infty} f(x) = L_-; \tag 6$$

i.e.,

$$\forall 0 < \epsilon_- \in \Bbb R \; \exists 0 > M_- \in \Bbb R, \; x < M_- \Longrightarrow \vert f(x) - L_- \vert < \epsilon_-, \tag 7$$

or

$$x < M_- \Longrightarrow L_- - \epsilon_- < f(x) < L_- + \epsilon_-; \tag 8$$

it follows that, letting

$$b = \min(L_- - \epsilon_-, L_+ -\epsilon_+), \; B = \max(L_- + \epsilon_-, L_+ + \epsilon_+) \tag 9$$

and

$$m = \max(-M_-, M_+), \tag{10}$$

that

$$\vert x \vert > m \Longrightarrow b < f(x) < B, \tag{11}$$

so $$f(x)$$ is bounded on the set $$(-\infty, m) \cup (m, \infty)$$; furthermore, since the closed interval $$[-m, m]$$ is compact and $$f(x)$$ is continuous, $$\vert f(x) \vert$$ is strictly bounded on this interval by some $$0 < \beta \in \Bbb R$$:

$$x \in [-m, m] \Longrightarrow -\beta < f(x) < \beta; \tag{12}$$

combining (11) and (12) shows that $$f(x)$$ is bounded on all of $$\Bbb R$$. Indeed, we have

$$\forall x \in \Bbb R, \; \vert f(x) \vert < \max(\vert b \vert, \vert B \vert, \beta). \tag{13}$$

1) Consider $$[0,\infty)$$.

$$\lim_{x \rightarrow \infty}f(x)=L:$$

For $$\epsilon >0$$ there is a $$M$$, real, positive, s.t.

for $$x > M$$ $$|f(x)-L| <\epsilon$$, i.e.

$$- \epsilon +L < f(x) < \epsilon +L$$.

The continuous function $$f$$ attains minimum and maximum on the compact interval $$[0,M]$$.

Hence $$f$$ is bounded on $$[0,\infty)$$.

2) Proceed likewise for $$(-\infty,0]$$.

1) and 2): f is bounded on $$\mathbb{R}$$.

Also cf.. comment of miraunpajaro