Does $f\in L^2(\Bbb R^n)$, with $f=g+h$ imply both $g$ and $h$ belongs to $L^2$?

Question

I have $f\in L^1(\Bbb R^n)\cap L^2(\Bbb R^n)$, with $f=g+h$ such that $g,h\in L^1(\Bbb R^n)$.

Is it true that both $g$ and $h$ stays in $L^2(\Bbb R^n)$?

I tried directly $$ \int|f|^2=\int|g|^2+\int|h|^2+2\int gh $$ so if by contradiction for example $g\notin L^2(\Bbb R^n)$, RHS diverges, so $f$ would not stay in $L^2(\Bbb R^n)$, which is absurd.

But this doesn't convince me, because I'm not taking in account the behaviour of $\int gh$.

Any suggestions? Maybe using Holder inequality?

Answer 1

Trivial counterexample: Let $g \in L^1 \setminus L^2$. Then $f = g - g = 0 \in L^2 \cap L^1$ but $g, h = -g \notin L^2$.

Edit:

In fact as Arthur mentioned in the comment, you can augment the last $g$ with any $L^1 \cap L^2$-function. Generally we have: Let $f = g + h \in L^1 \cap L^2$ and $g,h \in L^1 \setminus L^2$. Then we have $g - f \in L^1 \setminus L^2$ and obviously $(g - f) + h = 0$. Hence every counter example is of that form.