21Can any one make me understand the solution in A.Engel?
They wrote its obvious that f(1) is minimum.
By the same logic f(1)<f(2) and f(1)<f(2)<f(3)<f(4).....
1)Find all function from +ve reals to the same satisfying f(f(x)-x) = 2x for all x>0
2) f is a function from the set of natural numbers to the same. satisfying f(n+1)>f(f(n)) for all natural number n.
Prove that f(n) = n
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9 Answers
21solved the 2nd one..
We want to prove f(n)≥n
we have f(n+1)> f(f(n)) for n = 1,2,3...
So must we have f(1) the minimum.(cz others are greater than f of something)
So f(1)≥1
Using induction let f(p-1)≥(p-1)
Given f(p)≥f(f(p-1)+1 (i will later prove if m≥n , f(m)≥n)
so f(p)≥f(p-1) + 1 ≥p-1+1≥p
so we have got f(p)≥p for all p, so f(p+1)>f(f(p))≥f(p) hence f is increasing.
again we need f(t+1)≥ f(f(t))+1 , from here its clear that must we have f(t)≤t for all t.
therefore f(x) = x for all natural x.
i used if m≥n , f(m)≥n
proof.
We use induction on 'n'.(that's the only hint i took , on ''n'')
when n = 1, its true.
Let m≥n-1 and we have f(m)≥n-1, implying f(f(m))≥n-1 (thats where i was stuck a while;))
we have f(m+1)≥f(f(m))+1≥n and this completes the induction.
21
11Take f(x)-x=g(x)
f(g(x))=g(x)+g(g(x))
So the eqn becomes g(x)+g(g(x))=2x.
Now take an interval P where g(x)>x. Partititon it into Pi 's such that
\bigcup_i{}P_i=P
The partitioning has been done such that there exists some k s.t g is monotonic in Pk.
WLOG take g to be monotonically increasing in that sub-interval.
Let Pk=[a,b]
Now from the assumption g9a)>a, which gives g(g(a))<a. We also get g(a)>b
Similarly we have g(g(b))<b
Now in any subinterval of Pk g is monotonically increasing....so that gives g(g(x))<x for all x in [g(a),g(b)]...
So g(g(g(a)))<g(a) & g(g(a))<a, so adding the two we get 2g(a)<g(a)+a - that gives g(a)<a, a contradiction....thus g(a)=a and extending it to [a,b] we get g(x)=x. The other case for g(x) monotonically decreasing or for g(x)<x can be settled similarly...
That also gives f(x)=2x as the only function satisfying the given functional eqn.
well it is pretty obvious why f(1) is the min.
now comparing f(1)with f(2)
either f(1)=f(2) or f(1) <f(2)
but f(1)=f(2) implies f(1)>f(f(1)) which contradicts the choice of f(1)
21this much is obvious.
but f(2)<f(3)<f(4)... is not that obvious i guess?
are actually wat we do next:
consider the set of f(x) for all x>=2
even this set has a least value element which namely bcoms f(2)...(similar arg. as in case of f(1))
also f(1)<f(2)
now comparing f(2) and f(3)
either they r equal or f(2)<f(3)
if f(2)=f(3)
f(2)>f(f(2)) .........................................................(1)
now f(2)>=2
however (1) does not allow f(2)=2
hence f(2)>f(f(2)) whr f(2) also>3
but this contradicts the choice of f(2) in the newly defined set...
