# Thread: A metric space of a collection of bounded functions

1. ## A metric space of a collection of bounded functions

Let $\displaystyle B[a,b]$ denotes the collection of all bounded functions from a closed interval [a,b] into R. Let $\displaystyle \rho$ be a metric for $\displaystyle B[a,b]$ such that $\displaystyle \rho(f,g) = lub \{ |f(x) - g(x) | : x \in [a,b]\}$.

Let $\displaystyle C[a,b]$ denotes the collection of all continuous real-valued functions on [a,b] such that $\displaystyle C[a,b]$ is a closed subspace of $\displaystyle B[a,b]$.
Let $\displaystyle \rho'$ be a metric for $\displaystyle C[a,b]$ such that $\displaystyle \rho'(f,g) = lub \{|f(x) -g(x) |: x \in [a,b] \}$.

(a) Show that $\displaystyle C[a,b]$ is not a compact in $\displaystyle B[a,b]$.
(b) Show that $\displaystyle C[a,b]$ is no where dense in $\displaystyle B[a,b]$.

(c) A sequence $\displaystyle \{f_{n}\}_{n=1}^{\infty}$ in $\displaystyle B[a,b]$ converges to a member $\displaystyle f \in B[a,b]$ with respect to the metric $\displaystyle \rho$ if and only if $\displaystyle \{f_{n}\}_{n=1}^{\infty}$ converges to f uniformly.

2. Originally Posted by aliceinwonderland
Let $\displaystyle B[a,b]$ denotes the collection of all bounded functions from a closed interval [a,b] into R. Let $\displaystyle \rho$ be a metric for $\displaystyle B[a,b]$ such that $\displaystyle \rho(f,g) = lub \{ |f(x) - g(x) | : x \in [a,b]\}$.

Let $\displaystyle C[a,b]$ denotes the collection of all continuous real-valued functions on [a,b] such that $\displaystyle C[a,b]$ is a closed subspace of $\displaystyle B[a,b]$.
Let $\displaystyle \rho'$ be a metric for $\displaystyle C[a,b]$ such that $\displaystyle \rho'(f,g) = lub \{|f(x) -g(x) |: x \in [a,b] \}$.

(a) Show that $\displaystyle C[a,b]$ is not a compact in $\displaystyle B[a,b]$.
(b) Show that $\displaystyle C[a,b]$ is no where dense in $\displaystyle B[a,b]$.

(c) A sequence $\displaystyle \{f_{n}\}_{n=1}^{\infty}$ in $\displaystyle B[a,b]$ converges to a member $\displaystyle f \in B[a,b]$ with respect to the metric $\displaystyle \rho$ if and only if $\displaystyle \{f_{n}\}_{n=1}^{\infty}$ converges to f uniformly.
Question (c) shows an interesting case as the following proof indicates.

For the forward implication we want to prove $\displaystyle f_{n}$ converges uniformly to f.

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

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

Let now $\displaystyle n\geq k$ and xε[a,b]...............

Then $\displaystyle |f_{n}(x)-f(x)|\leq lub \{|f_{n}(x)-f(x)|:x\in [a,b]\}=\rho(f_{n},f)<\epsilon$.

Since $\displaystyle |f_{n}(x)-f(x)|$ is a member of the set {$\displaystyle |f_{n}(x)-f(x)|:x \in [a,b]$}.

Thus $\displaystyle f_{n}$ converges uniformly to f.

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

1) To prove that uniform convergence implies,that $\displaystyle f_{n}(x)$ converges to f w.r.t ρ metric (or sup metric) ,and then prove that fεB[a,b],e.i f is bounded.

2)To prove that uniform convergence implies that f is bounded and then prove that $\displaystyle f_{n}$ 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 $\displaystyle n\geq k$ and xε[a,b],then |$\displaystyle f_{n}(x)- f(x)$|<ε........for all n and x................................................. ...2

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

....if $\displaystyle n\geq k$ and xε[a,b],then |$\displaystyle f_{n}(x)-f(x)$|<1 for all n ,x................................................ ........................................3

Put now n=k+1,

Then |f(x)|=$\displaystyle |f(x)-f_{k+1}(x)+f_{k+1}(x)|\leq |f(x)-f_{k+1}(x)| + |f_{k+1}(x)|\leq 1+|f_{k+1}(x)|$.

Since $\displaystyle f_{k+1}$ is bounded ,there exists M>0 such that:

.............if xε[a,b] ,then $\displaystyle |f_{k+1}(x)|\leq M$..................

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

.......$\displaystyle |f(x)|\leq c= M + 1|$,implying that f is bounded.

And now to prove that $\displaystyle f_{n}$ 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 $\displaystyle n\geq k$ and xε[a,b],then $\displaystyle |f_{n}(x)-f(x)|$<ε/2.....for all n , x................................................. .............

Also let $\displaystyle n\geq k$.

Since ε/2 is an upper bound for the set {$\displaystyle |f_{n}(x)-f(x)|:x\in[a,b]$},then lub{$\displaystyle |f_{n}(x)-f(x)|:x\in[a,b]$}<ε/2<ε,

and thus $\displaystyle f_{n}$ converges w.r.t sup metric ρ.