Page 1 of 1

### The set of all polynomials

Posted: Wed May 17, 2017 2:12 pm
Is the set of all polynomials open in $$\displaystyle{\left(C([-1,1])\,,||\cdot||_{\infty}\right)}$$ ?

### Re: The set of all polynomials

Posted: Thu May 18, 2017 7:44 pm
Polynomials being a subspace of Banach Space $(C[-1,1],\lVert \cdot \rVert_{\infty})$, must have empty interior, otherwise it'd have to be the full space.

(One argument for showing it's not closed, Polynomials have a countable Hamel basis and Banach spaces of infinite dimension cannot have a countable Hamel basis).

### Re: The set of all polynomials

Posted: Thu May 18, 2017 8:47 pm
Thank you r9m.

Here is another idea.

Suppose that $$\displaystyle{\mathcal{P}}$$ (the set of polynomials) is open in $$\displaystyle{\left(C([-1,1]),||\cdot||_{\infty}\right)}$$.

Let $$\displaystyle{f(x)=x\,,x\in\left[-1,1\right]}$$ and then $$\displaystyle{f\in\mathcal{P}\subseteq C([-1,1])}$$.

There exists $$\displaystyle{\epsilon>0}$$ such that $$\displaystyle{B(f,\epsilon)\subseteq \mathcal{P}}$$.

Consider the continuous function $$\displaystyle{g(x)=f(x)+\dfrac{\epsilon\,x}{2\,(1+|x|)}\,,x\in\left[-1,1\right]}$$

Observe that $$\displaystyle{\forall\,x\in\left[-1,1\right]\,\,,|g(x)-f(x)|=\dfrac{\epsilon\,|x|}{2\,(1+|x|)}<\dfrac{\epsilon}{2}}$$

so, $$\displaystyle{||g-f||_{\infty}\leq \dfrac{\epsilon}{2}<\epsilon\implies g\in B(f,\epsilon)\implies g\in\mathcal{P}}$$

which is a contradiction since $$\displaystyle{g}$$ is not twice differentiable at $$\displaystyle{x=0}$$.