Question (c) shows an interesting case as the following proof indicates.

For the forward implication we want to prove converges uniformly to f.

Hence let ε>0.Since converges to f w.r.t ρ in B[a,b],there exists a kεN (k belonging to natural Nos) such that:

.......if ,then lub{| -f(x)|:xε[a,b]}<ε..........for all n................................................. ........1

Let now and xε[a,b]...............

Then .

Since is a member of the set { }.

Thus converges uniformly to f.

Now for the converse they are two ways to do the proof:

1) To prove that uniform convergence implies,that converges to f w.r.t ρ metric (or sup metric) ,and then provethat fεB[a,b],e.i f is bounded.

2)To prove that uniform convergence implies that f is boundedand then provethat converges uniformly to f w.r.t sup metric.

I will follow the 2nd way and it is open for discussion in the forum to decide which procedure is the right one.

Hence ,uniform convergence implies that for all ε>0 there exists a k integer such that:

............if and xε[a,b],then | |<ε........for all n and x................................................. ...2

Let now xε[a,b],put ε=1,then there exists an integer k such that :

....if and xε[a,b],then | |<1 for all n ,x................................................ ........................................3

Put now n=k+1,

Then |f(x)|= .

Since is bounded ,there exists M>0 such that:

.............if xε[a,b] ,then ..................

So there exists a c for all xε[a,b] such that :

....... ,implying that f is bounded.

And now to prove that converges to f w.r.t supmetric ρ.

So let ε>0 .Again due to uniform convergence there exists an integer k such that:

.........If and xε[a,b],then <ε/2.....for all n , x................................................. .............

Also let .

Since ε/2 is an upper bound for the set { },then lub{ }<ε/2<ε,

and thus converges w.r.t sup metric ρ.