Is the function sequence of the form $f_{n}(x) = (x - x^{2})^n$ pointwise convergent?

Question

I believe that the sequence converges pointwise, but not uniformly. My reasoning is as follows:

By the way, the functions $f_n$ are defined in $[0, 1]$.

1. if $x \in \{0, 1\}, f_n(x) = 0^n = 0$, so you could pick any $N \in \mathbb{Z^+}$ and the distance from the limit of $f(x)$ would be $0$ in both cases, and thus less than any $\epsilon > 0$.

2. However, if $x \in (0, 1)$, then the choice of $N$ is dependent on $x$ as well as the choice of $\epsilon$. Therefore, the sequence is pointwise convergent and not uniformly convergent as uniform convergence requires the choice of $N$ to be independent of $x$.

I determined the value of $N$ and it seems like the proof works out. However, all functions $f_n$ would have a maximum, so I believe uniform convergence may also be possible.

I'd appreciate any thoughts you have on my "proof".

Answer 1

Actually, the convergence is uniform. For each $x\in[0,1]$, $0\leqslant x-x^2\leqslant\frac14$, and therefore $0\leqslant f_n(x)\leqslant\frac1{4^n}$. So, if $\varepsilon>0$, you just take some $N\in\Bbb N$ such that $\frac1{4^n}<\varepsilon$ and then, for each $n\geqslant N$ and each $x\in[0,1]$, $0\leqslant f_n(x)<\varepsilon$.

But your proof of the fact that $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function is correct.

Answer 2

Your argument is fine, but the convergence is also uniform.

Suppose that $x\leq\frac12$. Then, crudely, $$ (x -x ^2)^n\leq x ^n\leq\frac1{2^n}. $$ And if $x\geq\frac12$, then $1-x\leq\frac12$ and $$ (x-x^2)^n=\big((1-x)-(1-x)^2)^n\leq\frac1{2^n}. $$

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Alternatively, you can use calculus to find that your functions have a maximum at $x=\frac12$ and so $(x-x^2)^n\leq \frac1{4^n}$.