Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having difficulties with solving following problem.

Find a linear functional in $C[0,1]^*$ such that cannot be extended to $L^p [0,1]$ Why such a linear functional cannot be extended?

I know I should choose a linear functional which is not bounded... But then I don't know how to proceed that such an unbounded functional actually cannot be extended.

share|cite|improve this question
What does the * mean on C[0,1]*? – John Nov 6 '12 at 15:36
@John dual space of C[0,1] – Detectives Nov 6 '12 at 16:13
up vote 4 down vote accepted

Try a pointwise evaluation, i.e. $\delta_0:f\mapsto f(0)$.

Edit: A few clarifications.

First, $\delta_0$ is well-defined; if you want to see $C[0,1]$ inside $L^p$ already, then every time an equivalence class contains a continuous function, such function is unique.

As for why the extension cannot exist: first, one needs to say clearly what it means "doesn't exist". For linear functions can always be extended between vectors spaces (this is just linear algebra). The point here is that one wants a continuous extension. But the norm in $L^p$ is a different one, and sequences that didn't had a limit now do. For instance, consider the functions $g_n(t)=(1-nt)1_{[0,1/n]}$. They are all continuous and all satisfy $g_n(0)=1$, so $\delta_0(g_n)=1$ for all $n$. But, in $L^p$, $g_n\to0$, and thus $\delta_0(\lim g_n)\ne\lim\delta_0(g_n)$. So $\delta_0$ cannot be continuous.

share|cite|improve this answer
Is that well-defined, though? Technically, $L^p$ doesn't contain functions but rather equalence classes of functions which only differ on a set of measure 0. At least of you want $L^p$ to be a normed space... – fgp Nov 6 '12 at 15:21
@fgp You have a well defined embedding $C([0,1]) \to L^p([0,1])$. And on $C[0,1]$ evaluation is well defined ... – martini Nov 6 '12 at 15:46
@martini if we say pointwise evaluation at 0 be $\delta_0 $ , you mean that $ ||\delta_0|| = \displaystyle sup_{||f||_{\inf} =1} |\delta_0(f)| $ can be arbitrary large, since {0} has zero measure? – Detectives Nov 6 '12 at 16:19
@martini but then how can we say about extension of $ \delta_0 $ is impossible? – Detectives Nov 6 '12 at 16:22
@martini Yeah, I don't doubt that your answer is essentially correct. But since the OP seems to be new to functional analysis, I figured it'd make sense to point out that there is a small technical difficulty. If this is homework, I think he'll be expected to note that difficulty and e.h. state that if you embedd $C[0,1]$ into $L^P[0,1]$, each equivalent class in the image will contain exactly one continous function, and that it's that representative which is evaluated at $0$. – fgp Nov 6 '12 at 17:11

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.