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

real-analysissequences-and-seriesuniform-convergence

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.

Best Answer

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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