3
$\begingroup$

Problem statement:

We have show that $L^q [a,b] \subseteq L^p [a,b]$, for $1 \leq p \leq q$. Show that the inclusion is proper by showing that there exists a function $f \in L^p [a,b]$, but $f \notin L^q [a,b]$. For simplicity, take $[a,b]$ to be $[0,1]$. Show that the inclusion is valid if the interval $[a,b]$ is unbounded by giving an example of a function $f \in L^q [1, \infty)$, but $f \notin L^p [1, \infty)$, where $1 \leq p \leq q$.

The $L^p$ space is the space of all functions satisfying $(\int f^{p})^\frac{1}{p} < \infty$. Even the first statement of the problem is tricky-- is there a reference online or in a textbook that explicitly shows this? It would probably be an easier problem if I have seen the first inclusion which the book does not show.

Also, I looked through the other questions which handled this problem in a general measure space, but we are just working with functions of one variable so we must solve from first principles. Do I proceed with Holder's inequality?

$\endgroup$
3
$\begingroup$

For the first statement, that $L^q[a,b] \subset L^p[a,b]$, $1 \leq p < q$ (if $p = q$ it's trivial), you just need to use Holder's inequality: Assume that $f \in L^q[a,b]$, then:

$$ \int_a^b |f|^p \leq \left[\int_a^b |f|^q\right]^{p/q}\cdot \left[\int_a^b 1\right]^{1-p/q} = (b-a)^{1-p/q}\left[\int_a^b |f|^q\right]^{p/q} $$ Thus, $f \in L^p[a,b]$ since $f \in L^q[a,b]$. Note that this works since $p < q$, it follows that $q/p > 1$ in Holder.

$\endgroup$
  • $\begingroup$ Do I then just choose the $p$ and $q$ such that Holder's inequality fails for the second part of the question? $\endgroup$ – Jared May 29 '13 at 19:08
  • $\begingroup$ No, I think you want to find an example. Try playing around with the powers of the simplest function you can think of whose integral might blow up on $[0,1]$ (e.g. 1/x). $\endgroup$ – al0 May 29 '13 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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