Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For some $n\in\mathbb{N^+}$, let $\mathbb{P}_n,\mathbb{P}_{n-1}$ be the metric space of polynomials of order $n$ and $n-1$ respectively, both with the metric: $$\displaystyle \sup_{x\in[0,1]}\{|P(x)-Q(x)|\}$$ I want to find an upper bond $L$, in expression of $\epsilon$, such that $$\forall P(x),Q(x)\in\mathbb{P}_n, d(P(x),Q(x))<\epsilon\;\Longrightarrow \sum_{k=0}^n |p_k-q_k|<L$$ where $p_k$, $q_k$ are coefficients before the $x^k$ term in respectively $P(x)$ and $Q(x)$.

Could someone provide a very simple proof for such an L? (I'm a beginner so I don't know many theories, please don't go too deep).

Further question:

This step comes up when I was trying to prove that the differentiation map

$D:\mathbb{P}_n \to \mathbb{P}_{n-1}$ is continuous using $\epsilon$, $\delta$ proof. Is there an easy way to show this map is continuous apart from using the method above?

Thank you so very much.

share|cite|improve this question
up vote 1 down vote accepted

Identify $\mathbb{P}_n$ with $\mathbb{R}^{n+1}$ through the function that assigns to each polynomial $P(x)=\sum_{k=0}^np_kx^k$ the vector $(p_k)_{k=0}^n$ of its coefficients. Then $$ \|P\|_\infty=\sup_{x\in[0,1]}|P(x)| $$ and $$ \|P\|_1=\sum_{k=1}^n|p_k| $$ define norms on $\mathbb{P}_n$. Since all norms on a finite dimensional vector space are equivalent, there exists a constant $C>0$ such that $$ \frac{1}{C}\,\|P\|_1\le\|P\|_\infty\le C\,\|P\|_1\quad\forall P\in\mathbb{P}_n. $$ You will have to work a little harder to find an explicit value of $C$ (which might depend on $n$.)

share|cite|improve this answer
Oh yes, I will work out the rest. Thank you. – mez Jan 25 '13 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.