I need to show that $L^1\cap L^2$ is dense in $L^p$ for $p\in (1,2)$
I think one-way is to construct a sequence of function possibly simple in the intersection converging to any arbitrary function in $L^p$.
Also, I know a fact that, for all $ f\in L^p $ we can decompose them into $f_1+f_2$ where $f_1 \in L^r ,f_2\in L^s$ such that $p\in (r,s)$. Can this be helpful in someway.
Any suggestions will be helpful!! Thanks
Let $f$ be a function in $ L^p$ and let $$ f_n\colon x\mapsto f(x)\mathbf{1}_{[-n,n]}(x)\mathbf{1}_{[-n,n] }\left(\left\lvert f(x)\right\rvert\right). $$ The point of the indicator functions is that in general, $f$ has no reason to belong to $L^1$ or to $L^2$; if $f$ was bounded we would know that $f$ would be also in $L^2$ because $f^2\leqslant \lvert f\rvert^p \lVert f\rVert_\infty^{2-p}$; if $f$ had bounded support, we would know that $f\in L^1$. Now $f_n$ has both features hence $f_n$ belongs to $L^1\cap L^2$ and by applying the dominated convergence theorem to the sequence $\left(\left\lvert f_n-f\right\rvert^p\right)_{n\geqslant 1}$, we can show that $f_n\to f$ in $L^p$.