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as the title states, which $f_n$ converge uniformly on $[0,1]$

(i) $f_n(x)=\frac{x}{1+nx}~~~~~$ (ii) $f_n(x)=\frac{nx}{e^{nx^2}}~~~~$ (iii)$f_n(x)=n^{\frac{1}{2}}x(1-x)^{n}$

there are another 3 more but these are the ones ive done so far and i want to make sure my work is correct.

definition of pointwise convergence is

$$\forall \epsilon > 0, \wedge x\in[0,1] ~ \exists N\in\mathbb{N}. \text{s.t for } n>N \Longrightarrow |f_n(x)-f(x)|\leq \epsilon$$

then uniform convergence is

$$\forall \epsilon > 0,\exists N\in \mathbb{N} \text{ s.t } \forall n>N \Longrightarrow \sup_{x\in[0,1]}|f_n(x)-f(x)| \leq \epsilon $$

so for each of the above i just need to make sure that they first converge pointwise. then see if they converge uniformly. (now it's been a while since i last did this so i may be rusty which is why i'm looking for confirmation that i'm doing it right)

first things first, for (i): $$\lim_{n \longrightarrow \infty}f_n(x)=\lim_{n \longrightarrow \infty}\frac{x}{1+nx}=\lim_{n \longrightarrow \infty}\frac{n}{n}\frac{x/n}{1/n+x}=0 \text{ as }n \rightarrow \infty $$ so the function converges pointwise to the zero function (yes?)

so if this is uniformly convergent the limit of the sup of this function should also converge to the zero function so $$\sup_{x\in [0,1]}|f_n(x)-f|=\sup_{x\in [0,1]}|f_n(x)|=\sup_{x\in [0,1]}|\frac{x}{1+nx}|=\frac{1}{1+n}=0 \text{ as } n\rightarrow \infty$$

so it does converge uniformly. (yes?)

i use similiar reasoning on (ii) to argue that also converges uniformly and on (iii) i find $$\lim_{n \longrightarrow \infty}f_n(x)=\lim_{n \longrightarrow \infty}n^{\frac{1}{2}}x(1-x)^{n}$$ for x = 0 or 1 we have $f_n(x)=0~\forall n$

which i believes suggests $$0 \leq f_n(x) \leq 0 $$ which suggests $$f_n(x)=0$$ and alternative arguement which i'm more confident about is the idea that $(1-x)^{n}\rightarrow 0$ faster than $x\sqrt{n} \rightarrow \infty$ which would give me the same limit of $0$

am i on the right tracks?

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With regard to (iii):

For $0 \leqslant x < 1$, we have

$$ 1- x = \frac{1-x}{1-x +x} = \frac{1}{1 +\frac{x}{1-x}},$$

and using Bernoulli's inequality ($(1+a)^n > na$) it follows that

$$f_n(x) = n^{1/2}x(1-x)^n = \frac{n^{1/2}x}{\left(1 +\frac{x}{1-x}\right)^n} < \frac{n^{1/2}x}{\frac{nx}{1-x}} = \frac{1-x}{n^{1/2}}< \frac{1}{n^{1/2}}.$$

Since $f_n(1) = 0$ we have $|f_n(x)| < n^{-1/2}$ for all $x \in [0,1]$ and the convergence $f_n(x) \to 0$ is uniform.

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Your first two arguments look fine to me. In the third example you cannot conclude that the function is identically zero from the fact that it is at the boundaries. The second idea is better, but you can just compute the derivative of $f$, find that it vanishes at $x=\frac{1}{1+n}$ and using $f''$ that this is a maximum, which goes to zero roughly as $n^{-1/2}$. Then you can fix some $\epsilon$ and check the condition for uniform convergence to the zero function (since $f \geq 0$ in $[0,1]$) by computing the right $n$.

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