Continuity and liimit of functions

jeff1evesque
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Homework Statement


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

Is f necessarily continuous?

Homework Equations


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

The Attempt at a Solution


Question: Does the fact of knowing
lim_{n \rightarrow \infty}f_n(x) exists for each x \in [0,1]. Denote the limit by f(x).
give us insight to declare that f_n converges to f uniformly- and thus satisfying Arzela-Ascoli's theorem?

Thanks,Jeffrey Levesque
 
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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?
 
Can someone provide some insight for me as to what the following means:

jeff1evesque said:
lim_{n \rightarrow \infty}f_n(x) exists for each x in [0,1]. Denote the limit by f(x).

And how I could use this fact to construct my justification for whether f is necessarily continuous?
 
f isn't necessarily continuous. Face it. It says fn converges at each point. That's not enough to prove f is continuous.
 
Dick is suggesting that you find a counter-example. Taking fn to be piecewise linear will suffice.
 
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.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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