If $u \in L^1(\mathbb{R^n})$ we define the Fourier transform of that function: $\hat u(\xi)=\int u(x) e^{-ix \cdot \xi}dx$. It is also true that if $u,v \in L^1$ then we have $\int\hat u v dx =\int u \hat v dx$ the proof is straightforward due to Fubini theorem.
Is the following statement true?
If $u \in L^p, v \in L^{p'}$ then $\int \hat u v dx=\int u \hat v dx$
I think this is not true because this integral may not be defined. With $p'$ I mean the Hölder conjugate of $p$.
It is true if $p=2$. If $u∈ L^p$ and $v∈L^p$ (not $L^{p'}$!) with $p<2$ then your equality is still true since $\hat{u}$ and $\hat{v}$ are in $L^{p'}$, so both integrals are bounded by Hölder's inequality.
If $u∈ L^p$ with $p>2$, then $\hat{u}$ is not a function but a distribution, so you are right, your integral does not always make sense. And this is actually the way the Fourier transform is defined for tempered distributions (which contains all $L^p$): $$ \langle \hat{u},v\rangle := \langle u,\hat{v}\rangle $$ where $v$ is a smooth function with fast decay.