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I need to study the uniform convergence of

$f_n(x) = n(x-1)e^{-nx}$

on the interval $[0,+\infty)$

I've shown that :

  • at $x =0$ $f_n(0)=-n \xrightarrow{} -\infty$

  • at $x =1$ $f_n(1)=0 \xrightarrow{} 0$

  • for every $x \neq 0$ we have that $f_n(x)\xrightarrow{} 0$ for n $\rightarrow$ + $\infty$

It remains to prove the uniform convergence on the interval $[0,+\infty)$. But at $0$ there isn't pointwise convergence and the function $f(x)=0$ is defined for every $x \neq 0$. So in $[0,\infty)$ there isn't uniform convergence?

Thanks in advance for any help.

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  • $\begingroup$ $f_n(0) = -n$ which is not bounded and hence not converging. $\endgroup$ – Will M. Jan 22 at 21:12
  • $\begingroup$ I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,\infty)$. $\endgroup$ – Mark Jan 22 at 21:13
  • $\begingroup$ indeed the next question of my exercise is studying the uniform convergence in $[1,\infty]$. However i have to say that there isn't uniform convergence in $[0,+\infty]$. $\endgroup$ – andrew Jan 22 at 21:20
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You are right, it does not converge uniformly on $[0,\infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,\infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $\displaystyle{\lim_{n\to\infty}\sup_{x>0}|f_n(x)|=0}$. But $\displaystyle{\sup_{x>0}|f_n(x)|\geq\lim_{x\to0^+}|f_n(x)|=n}$, hence the limit above is equal to $+\infty$. Therefore the sequence does not converge uniformly on $(0,\infty)$ either. Let's take some small $\varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[\varepsilon,\infty)$. We have that $\displaystyle{\sup_{x\geq\varepsilon}|f_n(x)|=|f_n(\varepsilon)|}$ (take $\varepsilon\leq1$, it is easy to verify it). Now it is $|f_n(\varepsilon)|\to0$, therefore the convergence is uniform on $[\varepsilon,\infty)$ for any $\varepsilon>0$, but not on $[0,\infty)$, nor $(0,\infty)$.

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  • $\begingroup$ Can you explain in what way $\sup_{x>0} |f_n(x)| \geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$. $\endgroup$ – Diger Jan 22 at 21:50
  • $\begingroup$ @Diger Take any sequence $(x_m)\subset (0,\infty)$ converging to $0$. Obviously $\sup_{x>0}|f_n(x)|\geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $m\to\infty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$. $\endgroup$ – JustDroppedIn Jan 22 at 21:58
  • $\begingroup$ Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/ $\endgroup$ – Diger Jan 22 at 22:08
  • $\begingroup$ @diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is $\endgroup$ – JustDroppedIn Jan 23 at 0:09

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