Help Needed Proving Implication for Linear Functional on Banach Space

cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
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:
Physics news on Phys.org
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.
 
@cbarker1 : Please use a double hash to wrap your math. Like in ##l^{-1}##, instead of a single one , like #l^{-1}#.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top