I'm interested in the definition of Fourier transform of a function $ f \in L^p(\mathbb{R}^d) $, where $ p \in [1, \infty] $ and $ d \in \mathbb{N} $ is the dimension.
The way I understand is that one treats $ f $ as a tempered distribution $ \Lambda_f $, where $$ \Lambda_f h = \int_{\mathbb{R}^d} f(x) h(x) \, \mathrm{d} x \, . $$ The Fourier transform on the tempered distributions space $ \mathcal{S}'(\mathbb{R}) $ is then given by $ (\mathcal{F} \Lambda)(h) = \Lambda (\mathcal{F} h) $. This allows us to calculate the Fourier transform of e.g. Dirac distribution, since $$ (\mathcal{F} \delta)(h) = \delta(\mathcal{F} h) = (\mathcal{F} h)(0) = \int_{\mathbb{R}^d} h(\xi) \exp(- 2 \pi i \langle 0, \xi\rangle) \, \mathrm{d} \xi = \int_{\mathbb{R}^d} h(\xi) \, \mathrm{d} \xi \, , $$ so in fact $ \mathcal{F} \delta = 1 $ (the constant function). However, let's suppose that I want to calculate the Fourier transform of the function $ f \equiv 1 $ which from what I understand is again the Dirac delta distribution. As far as I know, $ \mathcal{F} f = \mathcal{F} \Lambda_f $ (is this wrong?), but I cannot show that this in fact yields $ \mathcal{F} f = \delta $.
The question I suppose is where is my reasoning faulty and how to find the Fourier transform of $ f \equiv 1 $.
You’re correct that the Fourier transform of an $L^p$ function is the Fourier transform of the tempered distribution induced by that function.
The function $f=1$ induces the functional $\Lambda(h) = \int h(x)dx$, i.e. the integration functional, acting on Schwarz functions. By definition, the Fourier transform of $\Lambda$ is precisely the functional $\mathcal{F}\Lambda$ which acts on Schwarz functions via $\mathcal{F}\Lambda(h) = \Lambda(\mathcal{F}h)$. But $\Lambda(\mathcal{F}h) $ is the integral of $\mathcal{F}h$, which is in fact $(\mathcal{F}\mathcal{F} h)(0)$. However, applying the Fourier transform twice is just reflection $h(x)\mapsto h(-x)$ (possibly with some constants depending on your definition) but since we’re evaluating at zero, this yields $h(0)$.