Problem of Hahn-Banach theorem

Question

in $(l_\infty)'$ space there is a $f$ functional that $$\forall x \in c_{00};f(x)=0$$ $$f((1,1,1,...))=1$$ Provide these properties. I find this functional and showed by using Hahn-Banach theorem please help

Answer 1

You can use an approach similar to my answer here. Since $c_0$ is a closed subspace and $x=(1,1,1,\ldots) \in \ell^\infty \setminus c_0$, we can (by using Hahn-Banach) find a linear functional $\lambda\in (\ell^\infty / c_0)'$ such that \begin{align*} \lambda(x+c_0) = \|x+c_0\|_{\ell^\infty/c_0} \neq 0. \end{align*} Now since $q\colon \ell^\infty \to \ell^\infty /c_0, y\mapsto y+c_0$ is continuous, we get $g := \lambda\circ q \in (\ell^\infty)'$ with \begin{align*} g(x) \neq 0 \qquad \text{and}\qquad g|_{c_0} = 0. \end{align*} We choose $f := \frac{1}{g(x)} g \in (\ell^\infty)'$ and thus get the desired functional, because $c_{00}\subset c_0$.

Answer 2

$Y=\overline{c_{00}}$ ,$x=(1,1,...)$ and $V=Y \oplus span\{ {x} \}$ $f:V \to \Bbb{F} $ ,$v=y+a.x \in V$,$y\in Y , a \in \Bbb{F}$ $f(v)=a.dis(x,Y)=a.inf_{y \in Y} ||x-y||$ f is a functional. Also $x=(1,1,1,...) \notin Y $ so $dis(x,Y)\neq 0 $ let $f(x)=f(0.y+1.x)=1.dis(x,Y)=m>0$ There is a $F \in (l_\infty)'$that $f=F|_V$ from Hahn-Bannach Theorem.$g(t)=\frac{F(t)}{m}$ $g((1,1,...))=\frac{F(1,1,1,...)}{m}=\frac{m}{m}=1$ and $\forall y \in c_{00}; g(y)=0$ $$$$ Let see f is a lineer functional .Take $v.w∈V$then $v=y1+a.x,w=y2+b.x,y1,y2∈Ya,b∈F $and take $λ∈F$. $$f(λ.v+w)=λ.f(v)+f(w)(?)$$ $$f(λ.v+w)=f(λy1+\lambda.a.x+y2+b.x)=f([λy1+y2]+[λa+b].x)$$ and $ λy1+y2∈Y \ then \ ∃y3∈Y \ that \ :y3=λy1+y2$ so we can write $$f(y3+[λa+b].x)=[λa+b]+dis(x,Y)=λ.a.dis(x,Y)+b.dist(x,Y)=λ.f(v)+f(w)$$