If $1<p<\infty, p^{-1}+q^{-1}=1,$ and $f\in L^{1}(\mathbb{R}^d),$ show that there are $g\in L^{p}(\mathbb{R}^d)$ and $h\in L^{q}(\mathbb{R}^d)$ such that $f=gh.$
$\textbf{My Thoughts:}$
I define $g,h$ as follows: $$g(x)=\begin{cases}f(x)^{1/p}&\text{if }f(x)\geq0\\(-f(x))^{1/p}&\text{if }f(x)<0.\end{cases}$$ $$h(x)=\begin{cases}f(x)^{1/q}&\text{if }f(x)\geq0\\(-f(x))^{1/q}&\text{if }f(x)<0.\end{cases}.$$ Since $f\in L^{1}(\mathbb{R}^d),$ we have that $$\|g\|_{p}=\left(\int_{\mathbb{R}^d}\vert f\vert\right)^{1/p}<\infty,$$ and similarly for $h$. Therefore, $g\in L^{p}(\mathbb{R}^d)$ and $h\in L^{q}(\mathbb{R}^d).$
Finally, if $f(x)\geq 0,$ then $g(x)\cdot h(x)=f(x)^{1/p}\cdot f(x)^{1/q}=f(x)$ since $p^{-1}+q^{-1}=1.$ Similarly, if $f(x)<0$ then $g(x)\cdot h(x)=(-f(x))^{1/p}\cdot (-f(x))^{1/q}=f(x).$
Do you agree with the above proof? Any feedback is much appreciated.
Thank you for your time.