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?


One can use the following result, known as Müntz Szasz theorem, for which a proof can be found here.

If $p\geqslant 1$ and $\lambda_n$ is a sequence of distinct real numbers greater than $-\frac 1p$, then $\operatorname{Span}\{x^{\lambda_n},n\in\mathbb N\}$ is dense in $L^p$ if and only if $$ \sum_{n=0}^\infty\frac{\lambda_n+\frac 1p}{1+\left(\lambda_n+\frac 1p\right)^2}=+\infty.$$

One can choose $\lambda_n:=\frac 1{n^2}-\frac 1p$ for the first question, and similar ideas for the second one.

  • $\begingroup$ It could be also good to look for a simpler example. $\endgroup$ – Davide Giraudo Aug 31 '13 at 16:39
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    $\begingroup$ this is madness if such an approach was expected on qualifying exam $\endgroup$ – Norbert Aug 31 '13 at 17:09
  • $\begingroup$ @Norbert I agree. This version of Müntz-Szasz theorem is probably not mentioned in any textbook. $\endgroup$ – Davide Giraudo Aug 31 '13 at 17:14

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 $f \notin L^p$.

  • $\begingroup$ +1: it's an approach which needs only knowledge which can be found in a textbook. It's probably what was expected. $\endgroup$ – Davide Giraudo Sep 3 '13 at 17:12

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