Injectivity of Fourier transform on $L^1+L^2$

Question

We know we can define the Fourier transform $\mathcal F_{L^1}: L^1(\mathbb R^d) \to C_0(\mathbb R^d)$ via an integral, and from there take $L^2$-limits to get a unitary map $\mathcal F_{L^2}:L^2(\mathbb R^d)\to L^2(\mathbb R^d)$. From there, we can prove $\mathcal F_{L^1}$ and $\mathcal F_{L^2}$ agree on $L^1\cap L^2$, and hence can get a well-defined linear map $\cal F$ on $L^1+L^2$. We know (by unitarity from Plancherel) that $\mathcal F_{L^2}$ is injective, and also from a version of Fourier inversion on $L^1$ (Injectivity of Fourier transform between $L^1(\mathbb{R})$ and $C_0(\mathbb{R})$) that $\mathcal F_{L^1}$ is injective, but my question is whether or not $\cal F$ on $L^1+L^2$ is injective.

The obvious first step is to break $f\in L^1+L^2$ into $L^1,L^2$ pieces $f=f_1+f_2$, and then suppose $$0=\mathcal F(f) := \mathcal F_{L^1}(f_1)+\mathcal F_{L^2}(f_2) \implies \mathcal F_{L^2}(f_2) = \mathcal F_{L^1}(-f_1) =: h \in L^2 \cap C_0,$$ but from there, I don't know what to do.

Amazingly, I searched Google ("fourier transform injective L^1+L^2") and found no results at all on this topic! I only came across this: https://mathoverflow.net/questions/238692/fourier-transform-surjective-on-lp-mathbbrn-for-p-in-1-2, which had a comment saying that $\mathcal F_{L^p}$ for $p\in (1,2)$ is injective, which would be a corollary of the result I'm asking for. But even that result I can not find any results on when querying Google ("fourier transform injective on L^p")! EDIT: I suppose Fourier inversion for $f \in L^p(\mathbb R^n)$ and $\hat f \in L^1(\mathbb R^n)$ , where $1<p<2$ proves the assertion that $\mathcal F_{L^p}$ is injective for $p\in (1,2)$.

Answer 1

I suppose we could consider simultaneous approximations by Schwarz functions (Schwartz class function convergence in $L^1$ and $L^2$), i.e. $\{s_n\}\in \mathcal S(\mathbb R^d) \subseteq L^1\cap L^2$ converging to $h$ in $L^1$ and $L^2$. Then applying (both versions of) Fourier inversion and using $(2,2)$-boundedness of $\mathcal F_2^*$ (in fact isometry) and $(1,\infty)$-boundedness of $\mathcal F_1^*$ produces

$$\|\mathcal F_2^*(s_n) - f_2\|_{L^2} = \|\mathcal F_2^*(s_n) - \mathcal F_2^*(h)\|_{L^2} = \|s_n - h\|_{L^2} \to 0$$ and $$\|\mathcal F_1^*(s_n) - (-f_1)\|_{L^\infty} = \|\mathcal F_1^*(s_n) - \mathcal F_1^*(h)\|_{L^\infty} \leq \|s_n - h\|_{L^1} \to 0.$$ But $\mathcal F_1^*$ and $\mathcal F_2^*$ agree on $L^1 \cap L^2$, so we have shown that $t_n := \mathcal F_2^*(s_n) = \mathcal F_1^*(s_n)$ is a sequence of Schwarz functions that converges in $L^2$ and $L^\infty$ to $f_2$ and $-f_1$ respectively. Convergence in $L^2$ implies a subsequence converges pointwise a.e. (Does convergence in $L^p$ imply convergence almost everywhere?), so indeed some subsequence $t_{n_k}$ converges pointwise a.e. to both $f_2$ and $-f_1$, so $f_2=_{a.e.} -f_1$. Thus $f= f_1+f_2 = 0$ a.e. as desired.