Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a subspace of $L^1(\mu)$.Construct a bounded linear functional on $M$ such that there are two (hence infinite) different linear extensions preserving norm to $L^1(\mu)$.

Someone outlined me as follows:pick $\tilde{f}$ on $\bar{M}$(the closure of $M$) such that $\tilde{f}=lim_{n\to\infty}f_n$ such that $\parallel \tilde{f}\parallel_1=\parallel f_n\parallel_1$,where $f_n$ are bounded linear functional defined on $M$,then put $F(x)=\frac{n}{n+1}\tilde{f}(x)$.

But I have some questions about it :

(1)does an extension be in accordance with a functional or several functionals?

(2)why should we define an functional $\tilde{f}$ on $\bar{M}$ first? Or,can we define an extension just like this:$F(x)=\frac{n}{n+1}f(x)$ where $f$ is a bounded linear functional on $M$.

We have known,by the Hahn-Banach theorem,we can find such an $F$.

(3)There isn't requirement of $M$ being closed in the Hahn-Banach theorem,if $M$ is closed,can we get some special arguments or a simplier proof?

Maybe the three questions have something resembly,but I haven't find it explicitly.I need your kind help,please.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.