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Let $\mathscr{P}_{l}$ denote the vector space of all polynomials of degree $\leq l$, with the norm $$||p||=\int_{-\infty}^{+\infty}|p(x)|e^{-|x|}dx.$$ I'm supposed to prove that this space is complete with regard to the corresponding metric. Is there a simple way to do this? Here's what I've got so far. Let $(p_{n})_{n \in \mathbb{N}}$ be a Cauchy sequence in $\mathscr{P}_{l}$, so then we have, for some arbitraty $\varepsilon>0$, and a sufficiently large $n_{0}$, for all $m, n \geq n_{0}$, $$\int_{-\infty}^{+\infty}|p_{n}(x)-p_{m}(x)|e^{-|x|}dx<\varepsilon$$ holds.

I want to prove that the difference between two coefficients of the same degree for $p_{n}, p_{m}$ becomes small, use that $\mathbb{R}$ is complete, and then get a polynomial $p_{\infty}$ which is determined by the values that the coefficients of the sequence converge to. I tried using the inequality $$\int_{-\infty}^{+\infty}|f(x)|dx \geq |\int_{-\infty}^{+\infty}f(x)dx|,$$ but this only works on even degree coefficients. Is this the right approach? Is there an easier way to do this?

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    Note that $P_n(\mathbb R)$ is a finite dimensional vector space over $\mathbb R$ (as $\dim P_n(\mathbb R) = n+1)$. You can then apply similar reasoning as here. – Mark Schultz-Wu Feb 06 '17 at 00:29
  • I have a question, though; when two normed spaces are equivalent, which properties exactly does that equivalence make them "share"? Because topological equivalence does not imply uniform equivalence. That is, completeness isn't a topological property. So when two normed spaces are equivalent, which "type" of equivalence is it? – Matija Sreckovic Feb 06 '17 at 01:43

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