The following proposition is an important result about how continuous functions behave on closed real intervals. It turns out that once a function is continuous, there is always a constant such that it cannot be exceeded by the function values, no matter which function we will take. This works only for closed intervals.

# Proposition: Continuous Real Functions on Closed Intervals Take Maximum and Minimum Values within these Intervals

Let $$[a,b]$$ be a closed real interval and let $$f:[a,b]\to\mathbb R$$ be an arbitrary continuous real function. Then there are two numbers $p,q\in[a,b]$ with $$\begin{array}{rcl}f(p)&=&\max\{f(x):~x\in[a,b]\},\\f(q)&=&\min\{f(x):~x\in[a,b]\}.\end{array}$$

### Notes

• This proposition does not work if the interval is open.
• For instance, the function $f:(0,1]\to\mathbb R$, $f(x)=\frac 2x$ is continuous and it takes a minimum at $x=1$ (which is $2$), but does not take a maximum on the open interval $(0,1].$
• Another example is the function $g:(0,1)\to\mathbb R$, $g(x)=x.$ It neither takes a maximum nor a minimum for all values of $x\in(0,1)$.

Proofs: 1 Corollaries: 1

Chapters: 1
Explanations: 2
Proofs: 3

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983