Is $L^p$ dense in $L^p+L^{\infty}$?

Question

With $p \ge 1$, and norm $\left| \! \left| f \right| \! \right|_{(L^p+L^{\infty})(\mathbb{R}^n,\mathbb{R})} = \inf_{g \in L^p, h \in L^{\infty}, f =g+h} (\left| \! \left| g \right| \! \right|_{L^p} + \left| \! \left| h \right| \! \right|_{L^{\infty}})$. I'm quite sure it's not dense but I don't want to miss something. For instance $\sin$ cannot be approached in this norm by a function in $L^p$ in think.

Answer 1

Let $f(t)=1$. For any decomposition $f=g+h$ with $g\in L^p$ and $h\in L^\infty$, given $m\in\mathbb N$ we have $$ m(\{|g|\geq\tfrac1m\})=\int_{\{|g|\geq\tfrac1m\}} 1\,dm=m\int_{\{|g|\geq\tfrac1m\}} \frac1m\,dm\leq m\,\|g\|_p^p<\infty. $$ So $m(\{|g|<\tfrac1m\})=\infty$. On this set, $$ |h|=|1-g|\geq1-|g|>1-\tfrac1m. $$ Thus $$ \|h\|_\infty\geq1-\tfrac1m $$ for all $m$, showing that $\|h\|_\infty=1$.

Applying the above to $f-g=-g+1$ we have shown that, for any $g\in L^p$,
$$ \|f-g\|\geq\|h\|_\infty=1. $$ Thus $$\operatorname{dist}(f,L^p)=1.$$