For what range of $a\in\Bbb R$ is $f(x)=ae^{-ax}$ in the unit ball with respect to the uniform metric on $[0,1]$?
I.e when is $\|f\|_\infty\lt 1$
So far I see that for $a\ge 0$, $\|f\|_\infty=\sup_{x\in[0,1]}|ae^{-ax}|=\sup_{x\in[0,1]}ae^{-ax}=a\lt1$ if $a\lt 1$, so we have: that $f$ is in the unit ball for $0\le a\lt 1$.
Then I considered $a\lt 0\Rightarrow \|f\|_\infty=\sup_{x\in[0,1]}|ae^{-ax}|=\sup_{x\in[0,1]}-ae^{-ax}$
If we let $b=-a$ we seek $b\gt 0$ such that:
$\sup_{x\in[0,1]}be^{bx}\lt1$
Which is now where I am stuck, my assumption is that the value is maximised when $x=1$, which gives us:
$be^{b}\lt1$.
From looking around, the result references the W lambert function, but my issue is that this seems a little over complicated for something that was worth about $3$ marks in an exam paper (from a total of $90$ marks).
Thanks in advance for any help!