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

Let $C_0(\mathbb{R})$ be the space of $\mathbb{R}$ valued sequences converging to $0$. Let $l_n$ be a positive sequence in $\mathbb{R}$ such that $\sum\limits_{n=1}^\infty l_n=1$. We define $$ F:C_0(\mathbb{R})\to\mathbb{R}:x\mapsto\sum\limits_{n=1}^\infty x_n l_n. $$ Further, define $M = \{x \in C_0(\mathbb{R}) : F(x) = 0\}$.

We are supposed to prove:

1) for $x \in C_0(\mathbb{R})$ and $y \in M\setminus\{x\}$, we have $|F(x)|\leq \sup\limits_{n\in\mathbb{N}}|x_n-y_n|$.

2) for $x \in C_0(\mathbb{R})$ we have $\mathrm{dist}(x,M)\leq|F(x)|$.

Our teacher gave us a hint for the second part which is to consider $x - F(x)z$, where $z$ is the constant $1$ sequence. This is not in $M$ but can be approximated in a suitable way.

How does one tackle this problem?

Thank you.

share|cite|improve this question
Who writes $C_0(\mathbb{R})$ for the space of real sequences? That's usually the space of continuous functions vanishing at infinity. Far more common is the notation $c_0$ or maybe $C_0(\mathbb{N},\mathbb{R})$. – t.b. Jun 5 '12 at 9:27
up vote 1 down vote accepted

1) Consider $C_0(\mathbb{R})$ with $\sup$ norm. Then you can easily check that $F\in C_0(\mathbb{R})'$ and $\Vert F\Vert=1$. Consider arbitrary $x\in C_0(\mathbb{R})$ and $y\in M$, then $F(y)=0$ and $$ |F(x)|=|F(x)-F(y)|=|F(x-y)|\leq\Vert F\Vert\Vert x-y\Vert\leq\Vert x-y\Vert=\sup\limits_{n\in\mathbb{N}}|x_n-y_n|. $$

2) Fix $x\notin M$, then from linear algebra we know that $C_0(R)=\mathrm{Ker}(F)\oplus\mathrm{span}\{x\}=M\oplus\mathrm{span}\{x\}$. Hence each $z\in C_0(\mathbb{R})$ have representation $z=x-ty$ for some $t\in\mathbb{R}$ and $y\in M$. Cosequently, $$ \Vert F\Vert= \sup\limits_{z\in C_0(\mathbb{R})}\frac{|F(z)|}{\Vert z\Vert}= \sup\limits_{y\in M,t\in\mathbb{R}}\frac{|F(x-ty)|}{\Vert x-ty\Vert} $$ Since $y\in M$ we have $F(x-ty)=F(x)-tF(y)=F(x)$, hence $$ \Vert F\Vert= \sup\limits_{y\in M,t\in\mathbb{R}}\frac{|F(x)|}{\Vert x-ty\Vert}= \sup\limits_{\hat{y}\in M}\frac{|F(x)|}{\Vert x-\hat{y}\Vert}= \frac{|F(x)|}{\inf\limits_{\hat{y}\in M}\Vert x-\hat{y}\Vert}= \frac{|F(x)|}{\mathrm{dist}(x,M)} $$ Finally using that $\Vert F\Vert= 1$ we obtain $$ \mathrm{dist}(x,M)=\frac{|F(x)|}{\Vert F\Vert}=|F(x)| $$

share|cite|improve this answer

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.