Stone-Weierstrass, uniform convergence

[holomorphic]

Homework Statement

Show that there are continuous functions $$g:[-1,1]\to R$$ such that no sequence of polynomials $$Q_n$$ satisfies $$Q_n(x^2)\to g(x)$$ uniformly on [-1,1] as $$n\to\infty$$

The Attempt at a Solution

Suppose there is a sequence $$Q_n$$ such that $$Q_n(x^2)\to g(x)$$ uniformly for $$g(x)=x$$.

Then $$\forall \epsilon > 0 \forall x \in [-1,1] \exists N:(n\geq N\Rightarrow |Q_n(x^2) - g(x)| \leq \epsilon)$$

Take $$\epsilon = 1/2$$. Then $$\exists N_1, N_2 : ( n \geq max\{N_1,N_2\} \Rightarrow |Q_n(1^2) - g(1)| \leq 1/2$$ and $$|Q_n((-1)^2) - g(-1)| \leq 1/2)$$.

Then for $$n \geq max\{N_1,N_2\}$$ we have
$$1 = 1/2 + 1/2 \geq |Q_n(1^2) - g(1)| + |Q_n((-1)^2) - g(-1)|$$
$$=|Q_n(1) - g(1)| + |Q_n(1) - g(-1)|$$
$$\geq |g(1)-g(-1)| = |1+1| = 2$$, which is false. Therefore there is no such $$Q_n$$.

Does this solution make sense?

 

[holomorphic]

Then $$\forall \epsilon > 0 \forall x \in [-1,1] \exists N:(n\geq N\Rightarrow |Q_n(x^2) - g(x)| \leq \epsilon)$$

Take $$\epsilon = 1/2$$. Then $$\exists N_1, N_2 : ( n \geq max\{N_1,N_2\} \Rightarrow |Q_n(1^2) - g(1)| \leq 1/2$$ and $$|Q_n((-1)^2) - g(-1)| \leq 1/2)$$.

Then for $$n \geq max\{N_1,N_2\}$$ we have
$$1 = 1/2 + 1/2 \geq |Q_n(1^2) - g(1)| + |Q_n((-1)^2) - g(-1)|$$
$$=|Q_n(1) - g(1)| + |Q_n(1) - g(-1)|$$
$$\geq |g(1)-g(-1)| = |1+1| = 2$$, which is false. Therefore there is no such $$Q_n$$.

Oops. I used pointwise convergence instead of uniform. Doesn't change much but this should have read:
Then $$\forall \epsilon > 0 \exists N: \forall x \in [-1,1], (n\geq N\Rightarrow |Q_n(x^2) - g(x)| \leq \epsilon)$$

Take $$\epsilon = 1/2$$. Then $$\exists N : ( n \geq N \Rightarrow |Q_n(1^2) - g(1)| \leq 1/2$$ and $$|Q_n((-1)^2) - g(-1)| \leq 1/2)$$.

Then for $$n \geq N$$ we have
$$1 = 1/2 + 1/2 \geq |Q_n(1^2) - g(1)| + |Q_n((-1)^2) - g(-1)|$$
$$=|Q_n(1) - g(1)| + |Q_n(1) - g(-1)|$$
$$\geq |g(1)-g(-1)| = |1+1| = 2$$, which is false. Therefore there is no such $$Q_n$$.

 

[losiu99]

It certainly makes sense. I haven't gone through all the details, but the main idea is surely correct.