Here is an image of the question
Consider the sequence of functions $(f_n)_n$ defined on [0,1] by:
$f_n(x) = x^nLogx$ if $x≠ 0$ & $f_n(0)=0$
Prove that the sequence of functions $(f_n)_n$ converges uniformly on [0,1].
I first showed that it is pointwise convergent and then I tried to show that it's uniformly convergent by taking the derivative of a function that bounds the given sequence (in hope of finding a maximum that depends only on n) but I can't seem to move further because the function that is bounding the sequence is strictly increasing between 0 and 1. Here is my work in details.
Also, in the next part. I'm trying to show that $g_n(x)$ is not uniformly convergent on [0,1]. I followed the same steps and couldn't get anywhere. I think if I can solve the first part then I can solve this one. Here is my work in details.
Thanks for any help in advance
Note that $f_n$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with $f_n'(x) = (n\log{x}+1)x^{n-1}$ for $x\in (0,1)$. Hence $f_n$ has a unique stationary point at $x = e^{-1/n}$ and since $f_n(0) = f_n(1) = 0$ we conclude $|f_n(x)| \le |f_n(e^{-1/n})| = \frac{e^{-1}}{n}$ for all $x \in [0,1]$. Since $ \frac{e^{-1}}{n} \to 0$ as $n\to \infty$ we conclude $f_n$ converges uniformly to $0$.
Applying the above to $g_n = nf_n$ we see $g_n(e^{-1/n}) = e^{-1}$ so $g_n$ cannot converge uniformly to $0$ (and since it converges pointwise to $0$ it cannot converge uniformly to anything else either).