Inclusion of $L^p$ spaces, bounded or unbounded

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?

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.
• Do I then just choose the $p$ and $q$ such that Holder's inequality fails for the second part of the question? – Jared May 29 '13 at 19:08
• 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). – al0 May 29 '13 at 19:14