0
$\begingroup$

Show that for any $1\leq p<\infty$, the set $L^1\cap L^p$ is a dense subset of $L^p$.


Let $f\in L^p-L^1$. We need to find a sequence $\{\phi_n\}_n$ in $L^1\cap L^p$ converging to $f$. And I know the simple approximation theorem.

I think the following lemma is useful.

Lemma: If a simple function in a measure space $(X,\mathfrak{B},\mu)$ which belongs to $L^p(\mu)$, $1\leq p < \infty$, also belongs to $L^1(\mu)$.

Attempt: Let $g$ be simple function in $L^p$. Then we have $g=\Sigma_{i=1}^{m}a_i\chi_{E_i}$ for some $E_1,...,E_m\in\mathfrak{B}$ and some $a_1,...,a_m$ in $\mathbb{R}$. So $|g|^p=\Sigma_{i=1}^{m}|a_i|^p\chi_{E_i}$. Since $g \in L^p$, we have $\infty >\int|g|^pd\mu=\Sigma_{i=1}^{m}|a_i|^p\int \chi_{E_i}d\mu=\Sigma_{i=1}^{m}|a_i|^p\mu{E_i}$ . So $\mu(E_i)<\infty$ for all $i=1,...,m$. So $\infty>\Sigma_{i}^{m}|a_i|\mu(E_i)=\Sigma_{i=1}^{m}|a_i|\int \chi_{E_i}d\mu=\int|g|d\mu$. So $g \in L^1(\mu)$.

How is my attempt? How can we conclude the proof? Thanks!

$\endgroup$
0
$\begingroup$

It may be easier to work with truncations rather than simple functions. Given $f \in L^p$ define $f_n(x) = f(x)$ if $|f(x)| > \frac 1n$, and $0$ otherwise. Then $f_n(x) \to f(x)$ for all $x$. Since $|f_n| \le |f|$ you have that $f_n \in L^p$, and since $|f_n - f|^p \le 2^p |f|^p$ LDCT implies $$ \lim_{n \to \infty} \int |f_n -f|^p \, d\mu = 0. $$

On the other hand, assuming $p > 1$, you have by Holder's inequality and Chebyshev's inequality $$ \int |f_n| \, d\mu = \int_{\{|f| > \frac 1n\}} |f| \, d\mu \le \mu(\{|f| > \tfrac 1n\})^{1/p'} \|f\|_p < \infty$$ so that $f \in L^1$ too.

$\endgroup$
0
$\begingroup$

Your proof is correct! As for density, one usually argues that $C_c^{\infty}$ is dense AND contained in $L^p$ for all $p\in [1, \infty)$ and hence lies also in your intersection.

$\endgroup$
  • $\begingroup$ How is $C_c^\infty$ defined on an arbitrary space $X$? $\endgroup$ – Umberto P. Jan 21 '15 at 19:54
  • $\begingroup$ Haha good point. I retract my answer. $\endgroup$ – user180850 Jan 21 '15 at 20:13

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.