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Let $X$ be the vector space of all polynomials with real coefficients and let $(a_0, a_1, \ldots)$ be a sequence of positive real numbers. I was able to show that with the prescription $\|p\| = \max_{n \in \mathbb{N}} |a_n p^{(n)}(0)|$, a norm is defined on $X$, where $p^{(n)}$ is the nth derivative of the polynomial $p$.

(a) Is $(X, \|\cdot\|)$ Banach? I don't think so, so I need to find a sequence $\{p_n\}$ that is Cauchy but not convergent. Any ideas?

(b) Is the space $\{p \in X \,|\, p'(0) = p''(0) = 0\}$ a closed subspace of $X$? I know it's closed if it's actually a subspace (which it obviously is), and for any sequence in this space, there exists a limit in $X$ which is actually in this space. So how do I prove this, because I think it's closed.

  • Hints: Let $p_n$ be the first $n$ terms of the power series for the exponential. The maps $p \mapsto p^{(k)}(0)$ are continuous with respect to said norm. – copper.hat Jun 01 '24 at 23:05

2 Answers2

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Let $$p_m(x)=\sum_{k=1}^m{1\over ka_k}{x^k\over k!}$$ The polynomials $p_m$ form a Cauchy sequence as for $m_1<m_2$ we have $\|p_{m_2}-p_{m_1}\|={1\over m_1+1}.$ However the limit does not exist.

More generally any infinite dimensional normed space with countable Hamel basis is not complete see.

Concerning the second part, the linear functionals $p\mapsto p'(0)$ and $p\mapsto p''(0)$ are bounded, so the subspace is closed as the intersection of the kernels of these functionals.

Remark It can be shown that the space $X$ is isometrically isomorphic to the subspace of $c_0,$ consisting of the sequences with finitely many nonzero terms. The isometry is given by $x^n\mapsto a_nn!e_n$ where $e_n$ denotes the sequence with $1$ on the $n$th coordinate and $0$ on all other coordinates.

Ryszard Szwarc
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Let us define $b_n:=\max(1,a_n)$ for each $n\geq 0$ and the family of polynomials $p_n(x):=\sum_{k=0}^n \tfrac{1}{(k+1)!b_k} x^k$. Clearly, for any $m>n$ we have that $\|p_m-p_n\|\leq \tfrac{1}{n+1}$, so this sequence is Cauchy. However, the limit $\sum_{k=0}^\infty \tfrac{1}{(k+1)!b_k} x^k$ is not a polynomial and it does exist as a function since the series is dominated by the series defining the exponential. So $X$ is not Banach.

For the second part, note that if $p(t)=c_0+c_1 t+\dots c_m t^m$, then the condition $p'(0)=0$ simply means that $c_1=0$ and the condition on the second derivative means $c_2=0$. So the relevant subspace, call it $Y$, consists of all polynomials with constant term followed by degree $3$ term. Say you have a sequence of polys $\{y_n\}\subset Y$ such that they converge to some $y\in X$ w.r.t. your norm. If $y$ has a non-zero degree $1$ term, say $c_1\neq 0$, or a non-zero degree $2$ term, say $c_2\neq 0$, then the difference norm $\|y_n-y\|$ would be bounded below by $|a_1c_1|$ or by $|2a_2c_2|$, contradicting the limit. Therefore $y\in Y$ and $Y$ is closed.

AnCar
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