This is indeed a very good and natural question. We usually learn that there is a whole spectrum of $L^p$-spaces, but then in practice only $L^2$, $L^1$ and $L^\infty$ seem to pop up. Why should we care about $L^{3/2}$?
A possible answer comes from nonlinear analysis.
The basic observation is the fact that $$\lVert u^k\rVert_{L^p}=\lVert u\rVert_{L^{kp}}^k.$$ So, when dealing with nonlinear problems in $L^p$, we will have to play with $p$. We won't be able to stay in the same space all the time.
For an example of this, let us consider the following nonlinear PDE:
$$\tag{1}
-\Delta u (x)= u^2(x), \qquad x\in \mathbb{R}^3.
$$
The associated linear inhomogeneous problem
$$\tag{2}
-\Delta u= h
$$
can be solved satisfactorily via the Fourier transform, which works especially well in $L^2$. So, assuming that everything belongs to $L^2(\mathbb{R}^3)$ (let me be a bit sloppy here), we can Fourier transform termwise in (2) and write $\hat{u}(\xi)=\lvert\xi\rvert^{-2}\hat{h}$, which can then be anti-transformed back to
$$u(x)= \left(\lvert 4\pi y\rvert^{-1} \ast h\right) (x)\stackrel{\text{def}}{=}(-\Delta)^{-1} h.$$
(Note that $\lvert 4\pi y\rvert^{-1}$ is exactly the fundamental solution of the Laplace operator). Setting $h=u^2$, we can now reformulate the nonlinear equation (1) as follows:
$$\tag{3}
u=(-\Delta)^{-1}\left( u^2\right),$$
which is a fixed-point equation. We want to approach it via the contraction mapping principle, by showing that the mapping
$$\Phi(u)=(-\Delta)^{-1}\left( u^2\right)$$
is contractive on some complete metric space to be specified later - and this is where $L^{3/2}$ enters the picture.
We need some estimates on $\Phi$, and these are provided by the Hardy-Littlewood-Sobolev inequality, which, in our case ($\alpha=2,\ n=3$), takes the form
$$\lVert (-\Delta)^{-1} f\rVert_{L^q(\mathbb{R}^3)} \le C_p \lVert f\rVert_{L^p(\mathbb{R}^3)}, \qquad 2+\frac{3}{q}=\frac{3}{p}.
$$
(The condition on $p$ and $q$ can be recovered via the scaling argument, by observing that both sides of this inequality are homogeneous with respect to the scaling $f(x)\mapsto f(\lambda x)$, and therefore the degrees of homogeneity must match). With $f=u^2$ this inequality reads
$$\tag{4}
\lVert \Phi(u)\rVert_{L^q(\mathbb{R}^3)}\le C_p \lVert u\rVert_{L^{2p}(\mathbb{R}^3)}^2.$$
It is now clear that our hands are tied: the only way to get something meaningful is to have $q=2p$, which means that $q=\frac{3}{2}$. Thus the right functional setting for this problem is the space $L^\frac{3}{2}(\mathbb{R}^3)$.
Indeed, if we let $B_R\subset L^{\frac{3}{2}}(\mathbb{R}^3)$ denote the closed ball of radius $R$, we see from (4) that $\Phi(B_R)\subset B_R$ if $R< 1/C_{3/2}$. Moreover, combining (4) with the Hölder inequality,
$$
\begin{split}
\lVert \Phi(u)-\Phi(v)\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)}&\le C_{3/2} \lVert u^2-v^2\rVert_{L^{\frac{3}{4}}(\mathbb{R}^3)} \\
&=C_{3/2}\lVert (u+v)(u-v)\rVert_{L^{\frac{3}{4}}(\mathbb{R}^3)} \\
&\le C_{3/2} \lVert u+v\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)}\lVert u-v\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)} \\
&\le 2RC_{3/2}\lVert u-v\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)}.
\end{split}
$$
This means that the map $\Phi\colon B_R\to B_R$ is contractive if $R<\frac{1}{2 C_{3/2}}$.
Conclusion. We have proved that:
- there is a unique solution to (1) in the ball of radius $1/(2C_{3/2})$ of the space $L^\frac{3}{2}(\mathbb{R}^3)$. Since $u=0$ is one such solution, that must be the only one.
- the sequence $$\begin{cases} u_{n+1}=(-\Delta)^{-1}\left( u_n^2 \right) \\ \lVert u_0\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)} \le R \end{cases}$$ converges to $u=0$ in the topology of $L^{\frac{3}{2}}(\mathbb{R}^3)$ regardless of the initial condition $u_0$ provided that $R<1/(2C_{3/2})$.
Fact 2 also highlights the necessity of considering convergence issues in $L^p$-spaces with $p \ne 2$.
