Uniform convergence of sequence of differentiable functions

Question

Show that the sequence of differentiable functions $x^n/n$ in $[0,1]$ converges uniformly to a differentiable function $f$ in $[0,1]$. Also, show that the sequence $f'_n$ converges to a function $h$ in $[0,1]$, but $h(1)\neq f'(1).$

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My attempt: I have proved that $f_n$ converges point wise to $0$ in $[0,1]$. I.e. for $x\in[0,1]$, $$\lim_{n\to\infty} x^n/n=0.$$

Now, $||f_n-f||=\sup_{0\leq x\leq 1} |x^n/n-0|=\sup_{0\leq x\leq 1}x^n/n=1^n/n$.

Then, $$\lim_{n\to\infty} 1^n/n=0$$. So, it converges uniformly in said interval. Is this attempt correct?

Also, how can I show that $f'_n$ converges to $h$?

Answer 1

your first proof seems good (of course u need to start your proof by ,let $n$ be a positive integer and then u write the calculations of the norm and replace $1^{n}$ by 1). now, a general remark is that if a real number $\alpha$ satisfies $\lvert \alpha \rvert<1$,then $\lim_{n\to \infty } \alpha^{n}=0$,so for $0\leq x<1$,$f_{n}'(x)\to 0$, whereas for $x=1$,the sequence is constant and $f_{n}'(1)\to 1$,So $h(x)=0$ for $0\leq x<1$ and $h(x)=1$ for $x=1$ and this doesn't equal $f'(1)=0$

Answer 2

Yes, the attempt is correct.

And, since $f_n'(x)=x^{n-1}$, $(f_n')_ {n\in\Bbb N}$ converges pointwise to $$ \begin{array}{rccc}h\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ if }x=1.\end{cases}\end{array} $$ So, $h(1)\ne f'(1)$.