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.