Continuity and liimit of functions

[jeff1evesque]

Homework Statement

Suppose [tex]f_n : [0, 1]\rightarrow R[/tex] is continuous and lim[tex]_{n \rightarrow \infty}f_n(x)[/tex] exists for each x in [0,1]. Denote the limit by [tex]f(x)[/tex].

Is f necessarily continuous?

Homework Equations

We know by Arzela-Ascoli theorem:
If [tex]f_n: [a,b] \rightarrow R[/tex] is continuous, and [tex]f_n[/tex] converges to [tex]f [/tex]uniformly, then [tex]f[/tex] is continuous.

The Attempt at a Solution

Question: Does the fact of knowing

lim[tex]_{n \rightarrow \infty}f_n(x)[/tex] exists for each [tex]x \in [0,1][/tex]. Denote the limit by f(x).

give us insight to declare that [tex]f_n[/tex] converges to [tex]f[/tex] uniformly- and thus satisfying Arzela-Ascoli's theorem?

Thanks,Jeffrey Levesque

 

[Dick]

Well, no. They only gave you that the fn are continuous and converge pointwise. I think they want you find an example of a sequence of continuous functions that are pointwise convergent but don't have an continuous limit. Can you think of one?

 

[jeff1evesque]

Can someone provide some insight for me as to what the following means:

lim[tex]_{n \rightarrow \infty}f_n(x)[/tex] exists for each x in [0,1]. Denote the limit by [tex]f(x)[/tex].

And how I could use this fact to construct my justification for whether f is necessarily continuous?

 

[Dick]

f isn't necessarily continuous. Face it. It says fn converges at each point. That's not enough to prove f is continuous.

 

[HallsofIvy]

Dick is suggesting that you find a counter-example. Taking f_n to be piecewise linear will suffice.

 

[Landau]

The Arzela-Ascoli theorem asserts something about a sequence of equicontinuous functions. This has little to do with your question [or you have seen a different version of A-A].

Just construct a counter-example, i.e. a sequence of continuous functions (f_n)_n which converges pointwise to some discontinuous function. (Every book that introduces the concept of 'uniform convergence' will have such a counter-example, so you probably have encountered one already.)