4.9.E: Problems on the Darboux Property and Related Topics
( \newcommand{\kernel}{\mathrm{null}\,}\)
Prove Note 1.
Prove Note 3.
Prove continuity at 0 in Example (c).
Prove Theorem 1 for polygon-connected sets.
[Hint: If
B⊇m−1⋃i=0L[¯pi,¯pi+1]
with
f(¯p0)<c<f(¯pm),
show that for at least one i, either c=f(¯pi) or f(¯pi)<c<f(¯pi+1). Then replace B in the theorem by the convex segment L[¯pi,¯pi+1].]
Show that, if f is strictly increasing on B⊆E, then f−1 has the same property on f[B], and both are one to one; similarly for decreasing functions.
For functions on B=[a,b]⊂E1, Theorem 1 can be proved thusly: If
f(a)<c<f(b),
let
P={x∈B|f(x)<c}
and put r=supP.
Show that f(r) is neither greater nor less than c, and so necessarily f(r)=c.
[Hint: If f(r)<c, continuity at r implies that f(x)<c on some Gr(δ) (§2, Problem 7), contrary to r=supP. (Why?)]
Continuing Problem 4, prove Theorem 1 in all generality, as follows.
Define
g(t)=¯p+t(¯q−¯p),0≤t≤1.
Then g is continuous (by Theorem 3 in §3), and so is the composite function h=f∘g, on [0,1]. By Problem 4, with B=[0,1], there is a t∈(0,1) with h(t)=c. Put ¯r=g(t), and show that f(¯r)=c.
Show that every equation of odd degree, of the form
f(x)=n∑k=0akxk=0(n=2m−1,an≠0),
has at least one solution for x in E1.
[Hint: Show that f takes both negative and positive values as x→−∞ or x→+∞; thus by the Darboux property, f must also take the intermediate value 0 for some x∈E1.]
Prove that if the functions f:A→(0,+∞) and g:A→E1 are both continuous, so also is the function h:A→E1 given by
h(x)=f(x)g(x).
[Hint: See Example (c)].
Using Corollary 2 in §2, and limit properties of the exponential and log functions, prove the "shorthand" Theorems 11−16 of §4.
Find limx→+∞(1+1x)√x.
Similarly, find a new solution of Problem 27 in Chapter 3, §15, reducing it to Problem 26.
Show that if f:E1→E∗ has the Darboux property on B(e.g., if B is convex and f is relatively continuous on B) and if f is one to one on B, then f is necessarily strictly monotone on B.
Prove that if two real functions f,g are relatively continuous on [a,b] (a<b) and
f(x)g(x)>0 for x∈[a,b],
then the equation
(x−a)f(x)+(x−b)g(x)=0
has a solution between a and b; similarly for the equation
f(x)x−a+g(x)x−b=0(a,b∈E1).
Similarly, discuss the solutions of
2x−4+9x−1+1x−2=0.