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Proving that $g_n(x)=nxe^{-nx}$ does not converge uniformly

Function is defined on a compact set $E = [0,1]$

Maximum of $g_n(x)$ is $g_n(\frac{1}{n})=\frac{1}{e}$

let $\lim g_n(x) =g(x) = 0$

$\forall x$ Given $\epsilon >0$ $\exists N $ such that $$|g(x)-g_n(x)|< \epsilon$$ whenever $n ≥ N$

Now To be uniformly continuous this should work for all $x$ let $x= \frac{1}{n}$ Then since $\lim g_n(x) =g(x) = 0$

$\epsilon = \frac{1}{e^2}$ wouldn't work. So not uniformly continuous.

But doesn't this prove that $g_n(x)$ is not point wise convergent? I'm having trouble understanding.

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    Uniform convergence is defined on a set. So what is this set? Usually one studies it on compact sets $[a, b]$. – 2017-02-24
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    yes it is defined on a compact set $E = [0,1]$ – 2017-02-24
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    Then study the limit of the maximum. You'll have to make separate cases because $g_n$ is not monotone. – 2017-02-24
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    What is the definition of pointwise convergence? – 2017-02-24
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    The key issue here is notational. You set "$x=\frac{1}{n}$":note that $x$ is *not* the same for every $g_n$, so it does not "disprove" pointwise convergence. It'd be more correct to write $x_n=\frac{1}{n}$ to emphasize this dependence. For any **fixed** $x$, $g_n(x) \xrightarrow[n\to\infty]{}0$ (pointwise convergence); **but** there exists a sequence of points $(x_n)_n$ such that $$\lVert g_n - 0\rVert_\infty \geq \lvert g_n(x_n)\rvert \not\xrightarrow[n\to\infty]{}0$$ (so no uniform convergence). – 2017-02-24

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No, it just shows that for that $\varepsilon$, you cannot find big enough $n$ that will work for every point simultaneously. Pointwise convergence means once you have fixed $\varepsilon$ AND fixed a point, THEN you can find $n$ big enough to make $|f_n(x)-f(x)|<\varepsilon$. For each $x>0$, $\lim\limits_{n\to\infty}nxe^{-nx}=\lim\limits_{y\to\infty}\dfrac{y}{e^y}=0$.

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    fix $\epsilon $ and fix $x=\frac{1}{n}$ Now no matter what $n$ is, my function is equal to $\frac{1}{e}$ So changing $n$ won't matter. That's what I'm saying. So I couldn't get what you mean. Do you mean by setting $x= \frac{1}{n}$ I'm not actually fixing $x$ but instead changing it? And how can I prove that this function is point wise convergent on all points on where it's defined? – 2017-02-24
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    Read my comment below your question. You are not fixing a $x$ that works for all functions ($\exists x, \forall n, g_n(x)...$): you are, for each function, finding a $x$ ($\forall n, \exists x, g_n(x)\dots$). So this does not contradict pointwise convergence, which would be the former. – 2017-02-24
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    @Xenidia: It doesn't make sense to say you're fixing $x=\frac1n$ and then change $n$. You are not fixing anything, you're changing the point. Take a point $x$ and fix it. Now, maybe $x=1/n$ for some integer $n$, but that doesn't matter, except that it is a bad choice of notation, because we're using $n$ for the index of the sequence. The sequence of functions $g_m(x)$ converges to $0$ at $x$. $\lim\limits_{m\to\infty}g_m(1/n)=0$. You would have $g_n(1/n)=1/e$, but then consider $g_{n+1}(1/n)$, $g_{n+2}(1/n)$, $g_{n+3}(1/n)$, etc., and you get a sequence converging to $0$. – 2017-02-24