Convergence of a sequence in polynomials vector space

Let $N\in\mathbb{N}$ be a natural number and let $\alpha_0,\alpha_1,...,\alpha_N$ be real numbers such that $\alpha_i\neq\alpha_j \forall i\neq j$

We define in $X$ (the vector space of polynomials of degree $\leq N$) the following: $$\lVert\mathcal{P}\rVert = \sum_{k=0}^N\lvert \mathcal{P}(\alpha_k)\rvert, (\mathcal{P}\in X)$$ To prove:

1. $\lVert .\rVert$ is a norm.
2. The topology generated by this norm does not depend on the $\alpha_k$.
3. A sequence $\{\mathcal{P}_n\}$ in X converges with the norm $\lVert.\rVert_\infty$ ($\lVert f\rVert_\infty=sup\{\lvert f(t)\rvert: t\in[a,b]\}$) in $[a,b]$ if, and only if, there exists $\beta_0,\beta_1,...,\beta_N\in[a,b]$ with $\beta_i\neq\beta_j \forall i\neq j$ such that $\{\mathcal{P}_n(\beta_k)\}_{n\in\mathbb{N}}$ converges $\forall k=0,1,...N$.

The first and second points are trivial: proving that $\lVert.\rVert$ is a norm consists only in applying the definition and, since all the norms are equivalent in a finite dimensional vector space, the topology generated does not depend on the $\alpha_k$.

My problem reaches with the third point: do I have to use that the convergence of $\{\mathcal{P}_n\}$ is equivalent to the convergence of its coordinates?

Could you please give me any hint to solve this problem?

Thank you.

• The nomr you mention in (3) is not the norm you defined...is it? I mean, because you wrote $\;||.||_\infty\;$ ... – DonAntonio Feb 19 '14 at 16:55
• Yes, @DonAntonio, it is different. I edited the post explaining it, sorry. – Alejandro Feb 19 '14 at 16:59

I thought that I could use that, since all the norms are equivalent in any finite dimensional vector space, the convergence with the norm $\lVert.\rVert_\infty$ is equivalent to the convergence with the norm $\lVert.\rVert$, defined in 1.

This way, I could argue that the convergence with $\lVert.\rVert$ is equivalent to the convergence of every addend in the summation (if only one addend does not converges, the summation does not converges).

It's quite intuitive and I don't know if it is rigorous, what do you think?

• Any opinion? I am not completely sure of this. – Alejandro Feb 19 '14 at 17:47