# 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 $L^p(\mathbb{R})$. [closed]

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 BMay 5 '16 at 21:23

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• The set $\{u \in L^1 \mid \int u =0\}$ is closed in $L^1$. – Siminore May 5 '16 at 17:00
• Similar with math.stackexchange.com/questions/266049/… . – DuFong May 5 '16 at 19:22
• Did you mean $p>1$ for the first one? – zhw. May 5 '16 at 20:00
• How could that be the first one? It's the last one!! – zhw. May 5 '16 at 22:08
• sorry, in the first case p>1, for the second part of the problem p=1 – 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.
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.
• 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? – DuFong May 6 '16 at 1:01