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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.

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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
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1 Answer

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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 $$ \mathrm{dist}(x,M)=\frac{|F(x)|}{\Vert F\Vert}=|F(x)| $$

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