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I am new to MSE, so apologies for any inconveniences regarding my question. I am studying for the GRE's Math Subject and decided to go back and study certain important concepts from the Analysis series. I am reading Elementary Analysis by Ross, particularly section 24, Uniform Convergence.

We are shown the sequence of real valued functions $$f_n(x) = nx^n, \quad x \in [0,1),$$ and told that it does not converge uniformly on said interval. However, $\ f_n \rightarrow f$ point-wise on $[0,1)$ where $\ f(x) = 0, \ \forall x \in [0,1).$

This all makes sense and I am fine with it. However, it is later remarked that a sequence $\ (f_n)$ of functions on a set $\ S \in \mathbb{R}$ converges uniformly if and only if $$lim[sup\{|f(x)-f_n(x)|:x \in S\}] = 0.$$

This is what confuses me, because I am pretty sure (not $100 \%$ sure), that since $limsup = L = liminf$ for any convergent sequence (where $L$ is the limit), then for the example $f_n(x) = nx^n$, shouldn't this sequence of functions technically be uniformly convergent?

I'm most definitely missing something, though hopefully not anything glaringly obvious, and would appreciate some feedback. Thanks!

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  • $\begingroup$ I think my issue might be with how we calculate the limsup? We find the supremum by finding the max of the function (by setting $f_n'(x) = 0$) and then find the limit of $f_n$ at that max point? $\endgroup$ – Algebra 8 Dec 24 '17 at 4:47
  • $\begingroup$ This is not $limsup$ of a sequence. It is the limit of the sequence of supermums of the sequence of sets. $\endgroup$ – idok Dec 24 '17 at 5:05
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The thing is, for uniform convergence, the function sequence $(f_n)$ must converges for all $x$ in the interval. Pointwise convergence gives convergence of the sequence of numbers $(f_n(x))$, where $x$ is a pre-fixed point in the interval. When we compute $$\lim_{n\to\infty}\left[\sup\{|f(x)-f_n(x)|:x\in S\}\right]$$the point $x$ is not fixed, as the supremum of the limiting set can change (and in your example does change) as $n\to\infty$.

Pointwise convergence says that, for a fixed $x\in S$ we have $$\lim_{n\to\infty}|f(x)-f_n(x)|=0$$and in this case, we can use the result that says $\liminf=\lim=\limsup$. However, for uniform convergence we consider the limit of the supremum of the set $\{|f(x)-f_n(x)|:x\in S\}$, which is not the same as $\limsup |f(x)-f_n(x)|$ for a fixed $x$.

Essentially, this test for uniform convergence is seeing whether or not the maximum distance between the sequence functions and the pointwise limit function goes to zero. If this distance goes to zero, then the sequence functions approach the limit function uniformly, otherwise they do not. Pointwise convergence just shows that, at a specific point $x$, the distance between the value of the sequence functions and the value of the limit function at this point $x$ goes to zero. This does not guarantee that the maximum distance between the limit function and the sequence functions goes to zero, as the point $x$ that produces the maximum distance need not be the same point as $n\to\infty$.

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  • $\begingroup$ Thank you all for the wonderful answers. I believe that I see the mistake I was making; namely, that we are not looking the supremum of a number but that of an entire sequence of sets. So, I am looking at the supremum of the entire set of $x$'s in the interval. The thing is that I wasn't particularly focused on what the pointwise convergence was doing; I just assumed that if all the numbers do wind up going to $0$ then surely the supremum of the entire sequence of functions will end up going to zero. $\endgroup$ – Algebra 8 Dec 24 '17 at 6:13
  • $\begingroup$ However, after reading the answers you all gave me, it made sense that we're thinking of the sequence of the function in its entirety over the interval, not just some point. Therefore, the derivative of the function on the interval must be taken into consideration. The derivative of said function is unbounded with respect to $n$. Therefore, as $n \rightarrow \infty$, the derivative function of the sequence goes to infinity as well. That is interesting, because it seems that the function does not converge as whole to the point wise convergence. $\endgroup$ – Algebra 8 Dec 24 '17 at 6:16
  • $\begingroup$ If we look at the graph as $n$ increases, it seems like some points close to $1$ keep pushing the bounds of $f_n(x)$ up. Is it safe, then, to say that while the points of the function converge to zero, the function as a whole never totally converges to $0$? I know that this seems like repetition since I just said that, but I just want to make sure. Especially since this is such an eye-opening realization for me... $\endgroup$ – Algebra 8 Dec 24 '17 at 6:20
  • $\begingroup$ Pointwise convergence looks at the convergence of the function sequence at specific points, whereas uniform convergence is regarding the entire function as a whole. Indeed, as a whole (on the given interval) the functions $f_n$ do not converge to the zero function as $n\to\infty$, even though at any given point $x$, we have $f_n(x)\to 0$ as $n\to\infty$. $\endgroup$ – Dave Dec 24 '17 at 6:23
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For bounded functions $f:S\to \Bbb R$ and $f_n:S\to \Bbb R$ let $$\|f-f_n\|=\sup_{x\in S}|f(x)-f_n(x)|.$$ Consider $\|f-f_n\|$ to be a measurement of how closely $f_n$ approximates $f,$ not at any one $x\in S,$ but as an over-all measurement over all of $S.$ Consider $f$ and $f_n$ as single objects, and $\|f-f_n\|$ as the distance from $f$ to $f_n.$

The sequence $(f_n)_n$ converges uniformly iff $\|f-f_n\|\to 0$ as $n\to \infty.$

If $(f_n)_n$ converges uniformly to $f$ then for any $y\in S$ we have $$|f(y)-f_n(x)|\leq \|f-f_n\|,$$ so $f_n(y)\to f(y)$ as $n\to \infty.$

If $f_n(y)\to f(y)$ for each $y\in S$ we say that $f_n$ converges point-wise to $f.$ So uniform convergence implies point-wise convergence (but not vice-versa). If we are testing whether $(f_n)_n$ converges uniformly to some (any) $f$, we can, as a first step, identify the "candidate" $f$ by evaluating $\lim_{n\to \infty}f_n(y)$ for each $y\in S.$

With $S=[0,1)$ and $f_n(x)=nx^n,$ then $f_n(y)\to 0$ for each $y\in S.$ By the previous paragraph, if $(f_n)_n$ did converge uniformly to a function $f,$ then we would have $f(y)=\lim_{n\to \infty}f_n(x)=0$ for each $y\in S$..... But for any $n\in \Bbb N$ we can find some $y_n \in S$ with $y_n$ close enough to $1$ that $(y_n)^n>1/2.$ Then $f(y_n)>n/2$ and $f(y)=0$ so $$\|f-f_n\|\geq |f(y_n)-f_n(y_n)|=|0-f_n(y_n)|>n/2.$$ So the sequence $(f_n)_n$ in the Q is not a uniformly convergent sequence..

BTW . An important result that holds for all metric spaces $S, T$ (not just sub-spaces of $\Bbb R$ ), is that if $(f_n:S\to T)_{n\in \Bbb N}$ is a sequence of continuous functions converging uniformly to $f:S\to T$, then $f$ is continuous.

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  • $\begingroup$ Eye-opening answers, thank you all so much! $\endgroup$ – Algebra 8 Dec 26 '17 at 2:53
  • $\begingroup$ Merry Christmas $\endgroup$ – DanielWainfleet Dec 26 '17 at 3:32

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