How to deduce the sup of a function sequence $f_n$ without computing the derivative?

Question

I came across the following function sequence:

$$ f_n = e^{-nx}\sin(nx)$$

I'm asked if the following sequence converges uniformly on the set of positive real numbers $[0,\infty[$.

I found 2 different solutions, both correct and gives basically the same result, the first is quite logical but the second, even tho is logical too, I think I'm missing a trick.

Solution 1:

If we take $x_n = \pi/2n$ we will have that $m_n = f_n(\pi/2n) = e^{-\pi/2}$ and that obviously the sup of the function is bigger than $m_n$ so it's not converging to zero, thus it doesn't converge uniformly.

Solution 2 :

The 2nd solutions computes the derivative and find a long and cumbersome expression, then the author tells us that if we go in this way (computing $f'$ and analyzing it's variations) will take too much time and might lead to no where.

Instead, observe that: $f(1/n) = e^{-1}\sin(1)$, thus $|f_n - f| = |f_n| <= f(1/n) $ which doesn't converges to zero thus the function doesn't CU on $[0,\infty[$.

The 1st solutions seems obvious for me, I would choose an $x_n$ such that I can find a sequence independent from n and say that the sup of my function is always greater than this sequence which doesn't converge to zero.

However in the 2nd solution I can't understand how did the author deduce that $f(1/n)$ is the sup (greater or equal to $|f_n|$) without computing and checking the variation of $f_n'$.

Is there a trick in this case to say that $f_n$ is always smaller than $f(1/n)$? How can I deduce that in other examples?

Answer 1

Basically the author of the second solution did the exact same thing that you did: They realized that $f_n$ depends on $n$ and $x$ not really separately, but only via their product:

$$f_n(x) = f(nx),$$

where $f(x)=e^{-x}\sin(x)$. That means that for any constant $c$ you get $f_n(\frac{c}n)=f(c)$, so the function sequence $f_n$ cannot converge to the null function uniformly if $f$ isn't already the null function. You just have to pick a $c$ with $f(c) \neq 0$ to prove that.

You picked $c=\frac{\pi}2$, the author picked $c=1$. Both fullfil $f(c) \neq 0$, so both get the job done. Neither is inherently any better than the other, IMO.

Answer 2

The author did not deduce that $\sup\lvert f_n\rvert=e^{-1}\sin(1)$. What he or she deduced was that $\sup\lvert f_n\rvert\geqslant e^{-1}\sin(1)$. It follows from this that we don't have $\lim_{n\to\infty}\sup\lvert f_n\rvert=0$ and that therefore $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly to the null function. But it converges pointwise to that function. Therefore, $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly at all.