What is the Limit in a Function Space?

[jdcasey9]

Homework Statement

Let (X,d) and (Y,p) be metric spaces, and let f, fn: X -> Y with fn -> f uniformly on X. If each fn is continuous at xcX, and if xn -> x in X, prove that lim n-> infinity fn(xn) = f(x).

Homework Equations

llxll inf (the infinity norm of x) = max (lx1l,...,lxnl)

The Attempt at a Solution

fn and f are continuous, so f(x) is defined.

lfn(xn) - f(x)l <= llfn-fll inf (the infinity norm of fn-f) and llfn-fll inf -> 0 as n->infinity so lfn(xn)-f(x)l -> 0 as n -> infinity and lim fn(xn) = f(x) as n->infinity.

Seems very easy, so I probably am missing something.

 

[micromass]

Yeah, there's a small thing you've missed and I have the feeling you should have used the triangle inequality somewhere.. You say that

|f_n(x_n)-f(x)|\leq \|f_n-f\|_\infty

But this is'nt true. It is of course true that

|f_n(x)-f(x)|\leq \|f_n-f\|_\infty

So you just need to use the triangle inequality somehow...

 

[jdcasey9]

Alright,

lfn(xn) - f(x)l <= lfn(xn) - fn(x)l + lfn(x) - f(x)l <= lfn(xn) - fn(x)l + llfn - fll inf -> lfn(xn) - fn(x)) + 0 as n->inf.

So, now we need to show that lfn(xn) - fn(x)l -> 0.

Can we do this the same way?

lfn(xn) - fn(x)l <= lfn(xn) - f(xn)l + lf(xn) - fn(x)l -> lf(xn) - fn(x)l <= lf(xn) -f(x)l + lfn(x) - f(x)l -> lf(xn) - f(x)l -> 0 as n->inf

Can we say that lfn(xn) - f(xn)l -> llfn - fll inf norm? and that lf(xn) - f(x)l -> 0 as n-> inf? (without any further proof)

 

[micromass]

To prove that |f_n(x_n)-f_n(x)|\rightarrow 0.

We know that x_n\rightarrow x. Continuity of the f_n yields that f_n(x_n)\rightarrow f_n(x).

 

[jdcasey9]

Oh, ok, thanks I appreciate your help.