# Real analysis (Uniform continuous)

Let {fn} be a sequence of continuous functions, all defined on [a,b]. Suppose {fn(a)} is a diverging sequence of real numbers. Prove that {fn} does not converge uniformly on (a,b]. (Notice a is not included in this interval.)

Assume {fn} converges unifromly and then use the diverging sequence of real numbers as a contradiction to f being continuous in the end.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Proof:

Since is uniformly convergent on the interval , assume the limit function is , then for any , we can find some , such that for all , we have for all . Then for all , we have

Since are continuous function on , then we can find some , such that for any , we have and . Thus we have

So is a Cauchy sequence and thus must be convergent. However, the condition says that is a divergent sequence. We get a contradiction.

Therefore, is not uniformly convergent on the interval .

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