Prove or disprove that $W$ is a Banach space.

Question

True or false, and justify.

Let $W =\{ P: \textrm{polynomial of degree} \le100\}$ and $|| P || = \sum|a_n|$.

Then $|| P ||$ is a norm and $W$ is a Banach space under this norm.

I proved that $|| P ||$ is a norm, but how can I prove or disprove that $W$ is a Banach space?

I know that the set of all polynomials is not a Banach space, but in finite case it will be same?

Thanks a lot.

Answer 1

$\newcommand{\N}{\mathbb N}$ $\newcommand{\C}{\mathbb C}$ $\newcommand{\R}{\mathbb R}$ $\newcommand{\e}{\epsilon}$

As it has been pointed out in the comments you must specify over which ring of polynomials you are working. I suppose that you are working on $\mathbb C[x]$ or $\mathbb R[x]$ or over any other ring such that it is a Banach space. As you have proved that ||P|| is a norm we must see whether $W$ is a Banach space or not. Let's prove that $(W,||.||)$ is complete. Observe that as all the polynomials have the same maximum degree it means that the difference of them have the same maximum degree, that is $p_n -p_m \in W$. For that matter let $(p_n)_{n\in \N}$ be a Cauchy sequence in $W$. This means that for a certain $n_0 \in \N$ for all $n,m\geq n_0$ we have that $$||p_m-p_n||<\e.$$ This in turn can be rewritten as \begin{equation}\sum_{i=0}^{100} |a_{ni} -{a_{mi}}| < \e \end{equation} where $(a_{ni}), (a_{mi})$ are the coefficients of the polynomials $p_n$ and $p_m$ respectively. This means we have $101$ (what is important here is that it is a finite number) Cauchy sequences in $\C$ or in $\R$, and as they are complete they converge. That is a for a certain $n_1 \in \N$ we have that for all $n\geq n_1$ $$|a_{ni}-a_i|<\dfrac{\e}{101}$$ which combined with what we have seen before implies that for all $n\geq n_1$ \begin{equation} \sum_{i=0}^{100} |a_{ni} -{a_{i}}| < \e. \end{equation} Let's name $p = \sum_{i=0}^{100} a_i $. What we have seen before implies that $p_n \to p$. As $p \in W$ we have seen that W is a banach space.

This exercise illustrates that in this case the problem can be reduced to the study of the coefficients of the polynomial while the degree as long as it's bounded is not a problem. Remember that the set of the polynomials is not complete because the degree is not bounded and that helps you construct a Cauchy sequence that does not converge to a polynomial but to $e^x$, for example.

Answer 2

Take $\phi : (W,\left\lVert\cdot\right\rVert ) \to (\mathbb{R}^{101},\left\lVert\cdot\right\rVert_1) $ defined as $$ a_0 + a_1 x + \ldots a_{100} x^{100} \to \left(a_0,\ldots,a_{100} \right) $$

This is an isometric isomorphism therefore if $\left\{ p_n \right\}_{n\in\mathbb{N}}$ is Cauchy in $W$ we have $$ \epsilon > \left\lVert p_n - p_m \right\rVert = \left\lVert \phi(p_n) - \phi(p_m) \right\rVert_1 $$

So $\left\{ \phi(p_n) \right\}_{n\in \mathbb{N}} $ is cauchy as well so there's a $x \in \mathbb{R}^{101}$ such that

$$ \lim_{n\to\infty} \phi(p_n) = x \iff \lim_{n\to\infty} p_n = \phi^{-1}(x) $$ (Notice the two limits are w.r.t. the topology from $\left\lVert\cdot\right\rVert$ and $\left\lVert\cdot\right\rVert_1$ respectively).

Defining $p = \phi^{-1}(x) \in W$ we see that $p_n$ converges to $p$, hence $W$ is complete.