- June 24, 2019 at 6:38 pm #29369
Suppose that f is continuous on [a, b] and that f(a) < f(b). Prove that
there are numbers c and d with a <= c < d <= b such that f(c) = f(a)
and f(d) = f(b) and f(a) < f(x) < f(d) for all x in (c, d).
considering the following cases would be sufficient?
f is strictly increasing or increasing 0n (a,b)
f is decreasing in some interval and increasing in some interval?
How to give a better proof for this?June 24, 2019 at 11:10 pm #29380
A mind-blowing problem! This will be my hot favourite application of binary search algorithm and nested intervals theorem. Well, you can apply this technique in many other real analysis problems, so follow it carefully:
If c=a and d=b satisfy the conditions, then we are done. Otherwise, suppose there is an x in [a,b] such that f(x) is not between f(a) and f(b). Two cases arise:
(i) f(x)>f(b), in which case, by continuity there is a b1 in (a,x) such that f(b1)=f(b). In this case, let a1=a.
(ii) f(x)<f(a), in which case, by continuity again, there is an a1 in (x,b) such that f(a1)=f(a). In this case, let b1=b.
Repeat this process on a1 and b1 to get a2 and b2, and so on, till you get an interval, say [a(n),b(n)] for some n, such that f(a(n))=f(a), f(b(n))=f(b) and for all x in (a(n),b(n)), f(x) is in (f(a),f(b)). Letting c=a(n) and d=b(n), we are done.
Suppose the algorithm never stops, that is you never arrive at an interval [a(n),b(n)] satisfying the given conditions. Then by nested intervals theorem, there is a t in [a,b] such that a(n)–> t and b(n)—>t as n–>infinity. But by continuity, f(a(n))—>f(t) and f(b(n))–>f(t) . So f(a)—> f(t) and f(b)—->f(t) [ basically, f(a(n)) and f(b(n)) are constant sequences tending to the same limit.] So f(t)=f(a)=f(b), contradicting f(a)<f(b).
Still an amateur at analysis, so please point out any errors. Cheers!
- This reply was modified 3 weeks, 3 days ago by Alpha Beta.
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