# Uniform convergence of $f_n(x) = n(x-1)e^{-nx}$

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.

• $f_n(0) = -n$ which is not bounded and hence not converging. – Will M. Jan 22 at 21:12
• I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,\infty)$. – Mark Jan 22 at 21:13
• 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 $[0,+\infty]$. – andrew Jan 22 at 21:20

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 $$(0,\infty)$$.
• 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 $x=0$. – Diger Jan 22 at 21:50
• @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 $f_n$. – JustDroppedIn Jan 22 at 21:58
• 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 $x_m$ :-/ – Diger Jan 22 at 22:08