Help Needed Proving Implication for Linear Functional on Banach Space

[cbarker1]

Homework Statement

Show that for any linear functional $l$ on $B$ is continuous if and only if $A=\{f\in\B:l(f)=0\}$ is closed.

Relevant Equations

#B# is a Banach space over the complex field and #ker(l)={f\in \mcalB: l(f)=0}#

Dear everybody,

I am having some trouble proving the implication (or the forward direction.) Here is my work:

Suppose that we have an arbitrary linear functional $l$ on a Banach Space $B$ is continuous. Since $l$ is continuous linear functional on B, in other words, we want show that $l^{-1}\{0\}=A$ and this is closed. I am having trouble with this claim.

Thanks
Carter

 

[fresh_42]

Homework Statement:: Show that for any linear functional #l# on #B# is continuous if and only if #A=\{f\in\mclB:l(f)=0\}# is closed.
Relevant Equations:: #B# is a Banach space over the complex field and #ker(l)={f\in \mcalB: l(f)=0}#

Dear everybody,

I am having some trouble proving the implication (or the forward direction.) Here is my work:

Suppose that we have an arbitrary linear functional l on a Banach Space #B# is continuous. Since #l# is continuous linear functional on #B#, in other words, we want show that #l^{-1}{0}=A# and this is closed. I am having trouble with this claim.

Thanks
Carter

It is also equivalent to being bound. You can use this for the way back.

 

[WWGD]

@cbarker1 : Please use a double hash to wrap your math. Like in $l^{-1}$, instead of a single one , like #l^{-1}#.