prove that $f$ attains at least its minimum or maximum

Question

Let $f$ be a continuous function in the interval $[0,\infty]$. $\lim_{x\to\infty} f(x) = L.$ How would I prove that $f$ attains at least its minimum or maximum. I can use the definition of the limit to show that $f$ is bounded at $[N,\infty]$ by $ L+1$ and $L-1$ for example. And to use second Weierstrass theorem for a continuous funcion in a closed set and to conclude that it is bounded in $[0,N]$. Also it gets its minimum and maximum in that set. So, it is bounded in $[0,\infty]$. But I have a problem showing that it attains at least its maximum or minimum is $[0,\infty]$.

Thanks!

Answer 1

Let $N \geq 0$ be such that for all $x \geq N$, $|f(x) - L| \leq 1$. The extreme value theorem shows that $f$ has and attains both its maximum, $M_1$, and its minimum, $m_1$, on the compact set $[0,N]$.

Now consider $g(x) = f(1/x)$ for $1/x \geq N$, that is, $0 < x \leq 1/N$. From the continuity of $f$ on $[N,\infty)$, $g$ is continuous on $(0,1/N]$. Since $\lim_{x \rightarrow \infty} f(x) = L$ exists, we may continuously extend $g(x)$ to $x = 0^+$ by defining $g(0) = L$. (Equivalently, we may consider continuously extending $f$ to $[0,\infty]$.) $g$ is a continuous function on the compact set $[0,1/N]$, so the extreme value theorem shows it has and attains both is maximum, $M_2$ and its minimum, $m_2$ on $[0,1/N]$.

Let $m = \min \{m_1, m_2\}$, the global minimum of $f$ on $[0,\infty]$, and $M = \max \{M_1, M_2\}$, the global maximum of $f$ on $[0,\infty]$. If $m \neq M$, then at most one of $m$ or $M$ is $L$ so at least one of $m$ or $M$ is attained by $f$ on $[0,\infty)$. If $m = M$, then $f(x) = L = m = M$ for all $x$, so $f$ attains its maximum and minimum at every point of $[0,\infty)$.

Answer 2

HINT.-(1) Choose an interval $I_n=[n,n+h]$ contained in $[0,\infty]$ on which $f$ is not strictly increasing or decreasing; in this case by continuity an compacity we are done.

(2) If for all $I_n$ the function $f$ is strictly increasing or decreasing then $f(0)$ and $L$ are clearly extremums.

(3) If $f$ is increasing on $[n,n+h]$ and decreasing on $[n+h,n+h+k]$ then $f(n+h)$ is clearly a maximum and similarly when first decreasing and after increasing in which case $f(n+h)$ is a mim¡nimum.

Answer 3

(i). If $f(x)=L$ for all $x\in [0,\infty):$ We also have $f(\infty)=L.$ So $f(0)=\min f =\max f.$

(ii). If $x_0\in [0,\infty)$ with $|f(x_0)-L|=r\ne 0,$ take $x_1\in (x_0,\infty)$ such that $\forall x\in (x_1,\infty]\,(|f(x)-L|<r/2).$

Now $M=\max \{f(x):x\in [0,x_1]\}=f(x_2)$ exists for some $x_2\in [0,x_1],$ and $m=\min \{f(x):x\in [0,x_1]\}=f(x_3)$ exists for some $x_3\in [0,x_1].$

So if $f(x_0)>L$ then $f(x_2)=\max f,$ while if $f(x_0)<L$ then $f(x_3)=\min f.$

So in all cases $\exists y<\infty\,(f(y)=\max f \,\lor f(y)=\min f).$