Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have some difficulties to solve this problem

Let $f\colon\mathbb R\to\mathbb R$ a continuous function such that $$\lim_{x \to-\infty}f(x)=\lim_{x \to+\infty}f(x).$$ Prove that $ f $ has a minimum or a maximum in $ \mathbb R $ (in the sense that it could also have them both).

share|improve this question
2  
It is not hard, provided we know the relatively difficult result that a function continuous on a closed ordinary interval $[a,b]$ attains a max and a min on that interval. –  André Nicolas Jan 10 '12 at 8:52

3 Answers 3

up vote 4 down vote accepted

Write $\lim_{|x| \rightarrow \infty} f = l$. Fix $\epsilon$. By definition, away from an interval $I = [-R, R]$, $f$ lies within $\epsilon$ of $l$, and on $I$ $f$ is bounded by compactness, we deduce $f$ is bounded everywhere, hence $\inf f(\mathbb{R}), \sup f(\mathbb{R})$ exist.

From here it's case checking: if the $\inf$ and $\sup$ agree, $f$ is constant, if they disagree, $l$ is at most one of them, and since they are distinct they are separated by some epsilon, we deduce that the other must be obtained on an interval of the form $[-R, R]$ and by compactness is actually achieved by the function.

share|improve this answer

If $f$ is constant, then it has a maximum and minimum trivially. Suppose $f$ is not constant, so we have $f(a)-f(b)>\epsilon$ for some $\epsilon>0,a,b\in \mathbb{R}$. Since $\lim_{x \to-\infty}f(x)=\lim_{x \to+\infty}f(x)$ we have some $M$ such that $x\geq M\implies |f(x)-\lim\limits_{x\to-\infty} f(x)|<\epsilon/4$ and some $N$ such that $x\geq N\implies |f(-x)-\lim\limits_{x\to+\infty} f(x)|<\epsilon/4$, so for $|x|>\max\{M,N\}$ we have $|f(-x)-f(x)|<\epsilon/2$. If we choose $x$ such that $|x|\geq a,b$ then we have that $f$ must attain a minimum or maximum on $[-|x|,|x|]$, as we have some point in between which is either above or below both endpoints.

share|improve this answer

Suppose that the common limit is the real number $c$. Move $f$ up or down so that the limit is $0$. Call the resulting function $g$.

If $g$ is $0$ everywhere, we are finished. Otherwise, $g$ is either somewhere positive, or somewhere negative, or both.

Suppose first that $g$ is positive somewhere. Then for some $a$, and some positive $\epsilon$, we have $g(a)=\epsilon$.

There exists a real number $R$ such that $|g(x)|<\epsilon$ if $|x|>R$. On the interval $[-R, R]$, the function $g$ attains a maximum value. That value is at least $\epsilon$, so it is greater than $g(x)$ for any $x$ such that $|x| >R$. It follows that $g$ attains a global maximum.

If $g$ is somewhere negative, a mild modification of the above argument shows that $g$ attains a global minimum. Alternately, we can consider the function $-g$.

The result also holds if the limits are the same in the extended sense, for example if both limits are $+\infty$. Pick any real $a$, and let $f(a)=b$. There is an $R$ such that $f(x)>b$ if $|x|>R$. On the interval $[-R, R]$, the function $f$ attains a minimum value. That minimum value is $\le b$. Since $f(x)>b$ for $|x|>R$, that minimum value is a global minimum value.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.