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I was solving a question on uniformly convergent function. In my opinion, there is no point at which $nx(1-x)^n$ is discontinuous in [0,1]. Then it should be convergent in the interval [0,1].

Where am I going wrong with this?

Thanks in advance!

lumber jack
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    Converging at every point (pointwise convergence) is not the same as uniform convergence. I would check out this thread – Jspang Nov 28 '22 at 03:15
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    I suppose you are trying to apply Dini’s theorem. Then you should carefully read again how Dini’s theorem is stated. – Asigan Nov 28 '22 at 04:44
  • Thanks @Asigan for Dini's theorem. My book hasn't covered it till now. I had a preliminary glance at it right now. Will dig deeper into it. – lumber jack Nov 28 '22 at 07:41

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It is definitely possible for a sequence of functions to converge pointwise but not uniformly to a continuous function. A classic example of this is the sequence of triangle functions $f_n \colon [0, 1]\rightarrow \mathbf{R}$ defined by $$ f_n(x) = \begin{cases} nx & \text{if } 0 \leq x \leq 1/n, \\ 2-nx &\text{if } 1/n < x \leq 2/n. \\ 0 &\text{if } 2/n < x, \end{cases} $$ which converge pointwise to $0$ but $\sup_{x \in [0, 1]}|f_n(x) - 0| = 1$ for all $n$.

In your case, notice that \begin{align*} \sup_{x \in [0, 1]} |S_n(x) - 0| \geq S_n(1/n) = \frac{n}{n}\biggl(1 - \frac{1}{n}\biggr)^n \rightarrow e^{-1}, \end{align*} so that $$ \liminf_{n\rightarrow \infty} \sup_{x \in [0, 1]} |S_n(x) - 0| \geq e^{-1} $$ and so $\{S_n\}$ does not converge to $0$ uniformly.

Damian Pavlyshyn
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  • Hi @Damian Pavlyshyn, how do I know how to choose the correct $S_n(x)$ function to show the inequality(like you have chosen $S_n(1/n)$). This is very tricky IMO. Also, can you please share a link to the uniform convergence definition you have used? My book has provided only the standard definition of $|f_n(x)-f(x)|<\epsilon$ – lumber jack Nov 28 '22 at 03:29
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    There's no recipe for finding counter-examples, but a generic piece of advice is to plot functions to get some intuitive idea of how they behave. In this case, plugging in a few values of $n$ should be enough to see that $S_n$ has a peak whose location converges to $0$ as $n \rightarrow \infty$ but whose height does not vanish. Then you can do something like finding the maximum of $S_n$ (in this case, it happens at $1/(n+1)$) and understanding the function around there. In this case, the behaviour of $S(1/n)$ and $S(1/(n+1))$ is the same. – Damian Pavlyshyn Nov 28 '22 at 03:33
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    This is the standard definition of uniform convergence, ie $\sup_{x \in [0, 1]} |f_n(x) - f(x)| \rightarrow 0$. If you look closely in your book, I'm sure that you'll find that the definition is the same as this one, though it may be phrased as something more like: For all $\epsilon >0$, there exists $N_0$ such that $|f_n(x) - f(x)| < \epsilon$ for all $n > N_0$ and $x \in [0, 1]$. The key here is that it has to be the same $\epsilon$ for all $x$ simultaneously. – Damian Pavlyshyn Nov 28 '22 at 03:35