Show that for any $1<p<\infty$ the set $\{ f \in L^p(\mathbb{R}) \cap L^1(\mathbb{R})\}$ where $ \int_{\mathbb{R}} f=0$ is dense in $L^p(\mathbb{R})$. Is the statement true if $\mathbb{R}$ is replaced by $[0,1]$? Also what can we say when $p=1$?

Any clues? Thanks


closed as off-topic by Siminore, user296602, saz, user147263, John B May 5 '16 at 21:23

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  • $\begingroup$ The set $\{u \in L^1 \mid \int u =0\}$ is closed in $L^1$. $\endgroup$ – Siminore May 5 '16 at 17:00
  • $\begingroup$ Similar with math.stackexchange.com/questions/266049/… . $\endgroup$ – DuFong May 5 '16 at 19:22
  • $\begingroup$ Did you mean $p>1$ for the first one? $\endgroup$ – zhw. May 5 '16 at 20:00
  • $\begingroup$ How could that be the first one? It's the last one!! $\endgroup$ – zhw. May 5 '16 at 22:08
  • $\begingroup$ sorry, in the first case p>1, for the second part of the problem p=1 $\endgroup$ – Lucas May 5 '16 at 22:11

Lemma: Assume $p>1.$ Let $c\in \mathbb R.$ Then there exists a sequence $f_n\in L^1\cap L^p$ such that $\int f_n = c$ for all $n,$ and $\|f_n\|_p \to 0.$

Let's assume the lemma. Let $f\in L^1\cap L^p$ and put $c= - \int f.$ By the lemma, there are $f_n \in L^1\cap L^p$ such that $\int(f+f_n) = 0$ for all $n,$ with $\|f_n\|_p \to 0.$ Then $f+f_n \to f$ in $L^p,$ so we're done in the case $f\in L^1\cap L^p.$ Because $L^1\cap L^p$ is dense in $L^p,$ we've also proved the general result.

I'll leave the proof of the lemma to you for now.

Second question: Here we work on $[0,1].$ The answer is no. Let $f\equiv 1.$ Suppose there are $f_n\in L^1\cap L^p$ with $\int f_n = 0$ such that $f_n \to f$ in $L^p.$ Because convergence in $L^p$ implies convergence in $L^1$ on sets of finite measure (by Holder), we then have $0 = \int f_n \to \int f =1,$ contradiction.

  • $\begingroup$ Is the lemma still true if $\mathbb{R}$ replace by $[0,1]$? can we change variable like $y=tan(\pi/2(x-1/2)),x\in [0,1]$, then they become same problem? $\endgroup$ – DuFong May 6 '16 at 1:01
  • $\begingroup$ No, working on a set of finite measure changes things. See my answer to 2. above. $\endgroup$ – zhw. May 6 '16 at 1:20

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