Uniform convergence, wrong answer?

Question

I have the functions $$ f_n(x) = x + x^n(1 - x)^n $$

that $\to x$ as $n \to \infty $ (pointwise convergence).

Now I have to look whether the sequence converges uniformly, so I used the theorem and arrived at: $$ \sup\limits_{x\in[0,1]}|f_n(x)-x| = |x+x^n(1-x)^n - x| = |x^n(1-x)^n| = 0, n \to \infty $$

However, the master solution suggested something different that I'm unable to comprehend:

$$ \forall x \in [0,1]: 0 \leq x(1-x) = x - x^2 = 1/4 - (x - 1/2)^2 \leq 1/4 $$ $$ \sup\limits_{x\in[0,1]}|f_n(x) - x| = \sup\limits_{x\in[0,1]}|(x(1-x))^n| \leq (1/4)^n \to 0, (n \to \infty)$$

So, is my solution wrong? And how did they come up with that estimate? Or the better question is; how can I learn to come up with that stuff? First semester University and Analysis is... difficult.

Answer 1

When computing $\sup\limits_{x\in[0,1]}|f_n(x)-x|$ you should deduce something that does not depend on $x$.

Answer 2

Your idea seems perfectly fine. There are a lot of ways to go about the proof technically, but the method is very much the way corindo described it. M-test works, but there are a lot of ways to do this.

$(\forall x)(\forall \epsilon>0)(\exists N \in \mathbb{N})(n \geq N \implies |f_n(x)-f(x)|<\epsilon)$.

Let $\epsilon>0$. We know that $(\frac{1}{x})^{n}$ converges to $0$ for all $x \in [0,1)$, so just choose $N$ s.t $(\frac{1}{x})^{N}<\epsilon$. Since $(1-x)^n \leq 1$ for all $x$, the result follows readily.

For the case that $x=1$, the answer is almost trivial