Help Needed Proving Implication for Linear Functional on Banach Space

In summary, the conversation discusses the difficulty in proving the implication that any linear functional l on a Banach Space B is continuous if and only if the set A, defined as the kernel of l, is closed. The speaker is struggling with proving this claim and mentions using the concept of being bound as an alternative approach.
  • #1
cbarker1
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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
 
Last edited:
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  • #2
cbarker1 said:
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.
 
  • #3
@cbarker1 : Please use a double hash to wrap your math. Like in ##l^{-1}##, instead of a single one , like #l^{-1}#.
 

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