Define the functions $f_k : [0, 1] \to \mathbb{R}$ by $f_k(x) = k^ax \exp(−k^bx)$ where $a, b \gt 0$. How do I show that $f_k$ converges to $0$ uniformly if $a \lt b$ and that it does not converge uniformly if $a \geq b$? How do I find $||f_k||_\infty $?
For $a \lt b$ it seems quite intuitive as $\exp(-k^bx)$ $\to$ $0$ faster than $k^ax$ $\to$ $\infty$ however, how do I actually formalize this? Am I supposed to be using the M-test here?
We have $$ f_k'(x) = k^a \exp(-x k^b) (1 - x k^b). $$ This is zero when $x = 1/k^b$, and by checking the second derivative we see this is a maximum. Hence the largest $f_k(x)$ can be is $$ f_k(1/k^b) = k^{a - b} \exp(-1). $$ Now observe that if $a < b$, then the exponent in $k^{a - b}$ is negative, so the maximum of $f_k(x)$ is shrinking to zero in this case. Hence $f_k \to 0$ uniformly.
On the other hand if $a \geq b$, the exponent is nonnegative, whence the maximum of $f_k(x)$ goes to infinity (or possibly stays at $\exp(-1)$ if $a = b$), meaning the convergence here is not uniform.