TITLE: Subspaces of $L^p$
QUESTION [9 upvotes]: So studying Qualifying Exam problems in Analysis I cam across this one:
For $1\lt r \lt p \lt s \lt \infty$  where $\mu$ denotes Lebesgue measure,
a) Construct a subspace of $L^p([0,1],\mu)$  such that $ \forall r$ the subspace is dense in $L^r([0,1],\mu)$ but not $L^p$.   
b) Construct a subspace of $L^\infty([0,1,\mu)$ such that $ \forall s$ the subspace is dense in $L^p$ but not $L^s$.  
Now $L^s\subset L^p \subset L^r$ since $\mu([0,1])\lt\infty$, so the issue is recognizing the norms are not equivalent for this to be possible.  I'm just not terribly familiar with subspaces of $L^p$ for $p\neq2$.  Is this just a matter of better knowing $L^p$ spaces or is there something bigger that I'm missing?

REPLY [11 votes]: Here is a useful fact: if $X$ is a normed space and $\ell$ is a nonzero linear functional on $X$, then the kernel of $\ell$ is dense in $X$ iff $f$ is not continuous.
Here is another useful fact: given $\frac{1}{p'} + \frac{1}{q'} = 1$, we have
$$\|f\|_{p'} = \sup\left\{ \int fg : g \in L^\infty, \|g\|_{q'} = 1 \right\}.$$
(Use the fact that $(L^{q'})^* = L^{p'}$ and the fact that $L^\infty$ is dense in $L^{q'}$.)
Now let $\frac{1}{p} + \frac{1}{q}=1$, and choose $f \in L^q$ so that $f \notin L^{q'}$ for all $q' > q$.  (Consider a function of the form $f(x) = x^a (\log x)^b$.)  Now set
$$E := \left\{ g \in L^\infty : \int fg = 0 \right\}.$$
For $r < p$, let $t>q$ be the conjugate exponent of $r$.  Since $f \notin L^t$, the linear functional $g \mapsto \int fg$ is not continuous on $(L^r \cap L^\infty, \|\cdot\|_r)$.  Hence $E$ is dense in $(L^r \cap L^\infty, \|\cdot\|_r)$.  Since $L^r \cap L^\infty$ is dense in $(L^r, \|\cdot\|_r)$, we have that $E$ is dense in $(L^r, \|\cdot\|_r)$.
On the other hand, since $g \mapsto \int fg$ is continuous on $(L^p, \|\cdot\|_p)$, its kernel is closed.  $E$ is contained in that kernel and hence is not dense.
You can use a similar approach for part (b), constructing a function $f$ with $f \in L^r$ for all $r < p$ but $f \notin L^p$.