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For each $n = 1, 2, 3, \ldots$, let $f_n \colon [0,1] \to \mathbb{R}$ be defined by $$f_n(x) \colon= x^n \ \ \ \mbox{ for all } \ x \in [0,1].$$

Then $$ \lim_{n \to \infty} f_n(x) = \begin{cases} 0 \ & \mbox{ for } \ 0 \leq x < 1; \\ 1 \ & \mbox{ for } \ x=1. \end{cases} $$

So the point-wise limit of $f_n$ is a discontinuous function.

How to show directly that this sequence of functions does not converge uniformly?

Can we say that the only candidate for the uniform limit of $f_n$ is the function $f \colon [0,1] \to \mathbb{R}$ defined by $$ f(x) \colon= \begin{cases} 0 \ & \mbox{ for } \ 0 \leq x < 1; \\ 1 \ & \mbox{ for } \ x=1? \end{cases} $$

And if so, then can we say that since $f_n$ is a sequence of continuous functions and since $f$ is discontinuous (at $x=1$), therefore the convergence is not uniform?

Is it true that a uniform limit of a sequence of functions is always a point-wise limit?

And what about the converse?

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  • $\begingroup$ @Brian M. Scott, your answer is most eagerly awaited!! $\endgroup$ Apr 25 '15 at 11:22
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If the convergence were uniform, we would have $f_n(x)<1/3$ on $[0,1)$ for all sufficiently large $n.$ But note $f_n(1-1/n) = (1-1/n)^n \to 1/e > 1/3,$ contradiction

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