Not Difficult

Harry Potter

f(x) is a continuous function . We have 2 case :
1 ) $f: [a, + \infty ) - > R$ and $\lim_{x \to + \infty}{f(x)} = a$
2 ) $f: ( - \infty , + \infty ) - > R$ and $\lim_{x \to + \infty}{f(x)} = a$ , $\lim_{x \to - \infty}{f(x)} = b$

Prove that in all 2 cases , we have a Real number $M$ such that : $|f(x)| \leq M$

**Posted:** Wed Oct 21, 2009 1:33 am

mornik

There are many analogous cases in this proof, so I won't go into every detail - those analogies and the motivation for this solution seem obvious enough.

2) clearly follows from 1): divide $f$ into two parts: $f_1: \langle - \infty,a]\to\mathbb{R}$ and $f_2: [a, + \infty\rangle\to\mathbb{R}$ such that $f\mid_{\langle - \infty,a]} = f_1$ and $f\mid_{[a, + \infty\rangle} = f_2$ . Both functions are continuous. From 1), we'd get that $f_2$ is bounded and also, by analogy which will be obvious in the proof of 1), we get that $f_1$ is bounded too. Take the maximum of these two bounds and thus $f$ is bounded.

Now, about 1). Take $S = \{f(x): x\in[a,\ + \infty\rangle\}$ . Since $S$ is not empty, it has a supremum $t$ in $\overline{\mathbb{R}}$ . We want to prove that $t\in\mathbb{R}$ . In that case, $f$ has an upper bound. An analogous proof will work for $\inf S$ , so $f$ will have a lower bound too, and thus it will be bounded.

Right, so let's prove that $t\in\mathbb{R}$ . Let's suppose otherwise: $t = + \infty$ . Thus, there exists a sequence $(x_n)_{n\in\mathbb{N}}$ such that $f(x_n) > n$ for all $n\in\mathbb{N}$ . Since $x_n$ must have a limit point in $\overline{\mathbb{R}} = \mathbb{R}\cup\{-\infty, + \infty\}$ , take its subsequence $x_{p_n}$ such that $x_{p_n}\rightarrow L$ for some $L\in\overline{\mathbb{R}}$ . Obviously, $L\geq a$ so we don't have any problems with $f(L)$ . Now, if $L\in\mathbb{R}$ , by continuity we have $f(x_{p_n})\rightarrow f(L)$ . That is a contradiction since $f(x_{p_n}) > p_n > n$ for all $n\in\mathbb{N}$ . If $L = + \infty$ , we have $f(x_{p_n})\rightarrow A$ . However, that is again a contradiction (same reason: $f(x_{p_n}) > p_n > n$ ). (I don't know if you've made a typo or not, but you use $a$ in the sense of both $[a, + \infty\rangle$ and $f(x)\rightarrow a$ for $x\rightarrow + \infty$ . Of course, a stronger claim is still correct, we can have $\lim_{x\rightarrow + \infty}f(x) = A$ for any $A\in\mathbb{R}$ , not necessarily $A = a$ .)

**Posted:** Wed Oct 21, 2009 6:52 am

Harry Potter

Yes it is a foundation problem in analytics Smile

**Posted:** Thu Oct 22, 2009 12:38 am