Find the function

solved 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.

Take 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 P_i '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 P_k.

WLOG take g to be monotonically increasing in that sub-interval.
Let P_k=[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 P_k 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)

this much is obvious.

but f(2)<f(3)<f(4)... is not that obvious i guess?

why f(2) must be the second minimum?

for the first one,
f(x) = 0 and f(x) = 2x