inequality-why is it like that?

Question

I saw the solution of an exercise and there it is used the following inequality: $$e^{-(n-1)x} \leq e^{-{(n-1)}} ,\forall x \in [0,+\infty)$$

Why is it like that? I haven't understood it.. $$$$ The exercise is: Let $f_n:[0,+\infty) \to \mathbb{R}, f_n(x)=xe^{-nx}, \forall x\geq 0$.Prove that $(f_n)$ converges to a $f$ uniformly at $[0,+\infty)$.Which is the $f$?

This is the solution I found:

$f_n$ converges pointwise to $f(x)=0$.So,if it would converge uniformly,it would converge to $0$.We have $|f_n(x)-f(x)|=|f_n(x)|=|xe^{-nx}|=xe^{-x}e^{-(n-1)x} (1)$

For $\epsilon=1$ $\exists \delta>0$ such that : $\forall x> \delta$ we have $xe^{-x}<1$ In $[0, \delta], xe^{-x}$ is bounded as it is continuous in a closed and bounded interval. In $(\delta,+\infty)$,we have $xe^{-x}<1$.So, $xe^{-x}$ is bounded in $[0,+\infty)$, $xe^{-x} \leq M$

From the relation $(1)$ we have

$|f_n(x)-f(x)| \leq Me^{-(n-1)x} \leq Me^{-(n-1)}$

Taking the supremum,we have: $sup_{x \geq 0} {|f_n(x)-f(x)|} \geq Me^{-(n-1)} \to 0$.

So,it converges uniformly to $f=0$ in $[0,+\infty)$

It's good you don't understand it, because it's wrong. Is it perhaps for $x\in [1,\infty)$? — Ted Shifrin, Apr 06 '14 at 18:36
I will post as an answer the solution of the exercise,so that you tell me if it is wrong.. — evinda, Apr 06 '14 at 18:38
Don't post it as an answer. Edit your original post, please. — Ted Shifrin, Apr 06 '14 at 18:40

Ted Shifrin · Accepted Answer · 2014-04-06T20:28:52.270

3

I have no idea where you found this "solution." We want to find the maximum of $|f_n-f| = f_n$. The critical point of $f_n$ is $x=1/n$, so the maximum value of $f_n$ is $$M_n=f_n\left(\frac1n\right) = \frac1{ne}.$$ Since $M_n\to 0$ as $n\to\infty$, the convergence is uniform.

edited Apr 06 '14 at 20:28

answered Apr 06 '14 at 18:55

Ted Shifrin

115,160

I understand..thank you!!! – evinda Apr 06 '14 at 20:56

inequality-why is it like that?

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