Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]

Question

I can easily show that the set of polynomials on [0, 1/2] is not closed in C[0, 1/2] with the next example $$ g(x) = \delta \sum_{n=1}^{\infty}\frac{1}{n}x^n, x\in[0, 1/2] \\ g(x) = -\delta ln(1-x), x\in[0, 1/2] $$

Also, because $g(x) < \delta, x\in[0, 1/2]$, then $$ g(x) \in B(0, \delta) $$ , where $B$ is an open ball centered at $0$ with radius $\delta$.

My question is, how can I use this ball $B$ to show that my set of polynomials is not open either in C[0, 1/2]?

EDIT Probably I have to define my question better. What does it mean that my set is not open in C[0, 1/2]?

I have forgotten to say that I am using the infinity norm.

Davide Giraudo · Accepted Answer · 2013-05-08T13:44:24.057

0

The ball $B(0,\delta)$ is centered at a polynomial, but what is proved is that this ball contains $g$, which is not a polynomial.

In any metric space $(X,d)$, if we show that for a set $S$, there is a point $x_0$ for which the ball of center $x_0$ and radius $r$ is never contained in $S$, then $S$ cannot be open.

edited May 08 '13 at 13:44

answered May 08 '13 at 12:28

Davide Giraudo

172,925

And why is this an answer related to my question? I can not see it sorry. – May 08 '13 at 12:57
1

This proves that there is no $r$ such that $B(0,r)$ is contained in the space of polynomials. Actually, a vector subspace of a normed space which has a non-empty interior is the whole space. – Davide Giraudo May 08 '13 at 12:59
Now I see it, thank you very much. – May 08 '13 at 14:25
You are welcome. – Davide Giraudo May 08 '13 at 14:25

score 0 · Answer 2 · answered May 08 '13 at 12:33

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Hint: by linearity, all you have to prove is that for each $\varepsilon > 0$ there is a non-polynomial function $f$ with $\|f\| < \varepsilon$. Do you know any candidates?

answered May 08 '13 at 12:33

Aleksei Averchenko

8,023

I can find this candidate ($\delta = \epsilon$ in $g(x)$), but I do not see why is this related to my question, sorry. – May 08 '13 at 12:57
@noether what is the base of your topology? – Aleksei Averchenko May 08 '13 at 13:34

score 0 · Answer 3 · answered May 08 '13 at 14:00

Let $f$ be a polynomial and $\delta>0$ be given. Then $g(x):=f(x)+\delta\sin x$ has the properties

$g$ is continuos, so is on $C[0,\frac12]$
$\lVert g-f\rVert<\delta$, so $g$ is close to $f$, n fact $g\in B(f,\delta)$
For all $n>0$, the $n$th derivative is non-zero (always contains som $\sin$ or $\cos$), so $g$ is not a polynomial.

Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]

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