What's so special about $p=2$ for the $L^p$ spaces? The Banach space dual of $L^p$ is $L^q$, where $q=\frac{p}{p-1}$, but I don't really understand the motivation behind this. In particular, I find it kind of surprising that the only $L^p$ space whose dual is isomorphic $L^p$ is for $p=2$. So I guess I'm wondering what's so special about the number 2 in the context of $L^p$ spaces, or rather, where the formula $\frac 1 p + \frac 1 q = 1$ originally comes from / is motivated from.
Edit: I understand that $p=2$ gives the only Hilbert space, but I'm wondering whether there's any sort of deeper reason behind the relationship between $p$ and $q$ --- does $\frac 1 p + \frac 1 q = 1$ arise naturally out of integration theory in a more satisfactory way than "it just happens to be like that"?
 A: I elaborate on a remark in one of the other answers.
Consider a measure space $(X,\mathcal{M},\mu)$ which has sets of arbitrarily small measure and sets of arbitrarily large measure. (Here I think of the Lebesgue measure on the real line.) Let $1 \leq q < \infty$ and choose 
$$g \in L^q \setminus \bigcup_{p \in [1,\infty), p \neq q} L^p.$$
This is messy, but doable. $f \in L^1$, but $f \not\in L^p$ for all $p > 1$ handles the case $q=1$, and given my assumption about the measure space, you can do something similar to get higher $q$. Let $\phi(f) = \int_X fg d \mu$.
Now let us consider a function in $L^p$ which makes $\phi(f)$ "large". Specifically we take $f=|g|^{q/p} \text{sign}(g)$. The point is that this is automatically in $L^p$ since $g \in L^q$, but it is also big exactly where $g$ is big. 
Now if $\frac{q}{p} + 1 = q$ then $\phi(f)=\| g \|^q_{L^q}$. Yet otherwise $\phi(f)$ is infinite! That is, $\phi$ is only bounded on $L^p$ if $\frac{1}{p}+\frac{1}{q}=1$.
A: To expand a bit on the standard answer ("$p=2$ is a Hilbert space and Hilbert spaces are nice"), let's talk about why Hilbert spaces are nice.  One of the key feature of Hilbert spaces is that they are self dual.  This is expressed through the Riesz representation theorem, which says that if I want to define a bounded, linear map $\phi:L^2\rightarrow \Bbb{C}$, I just have to find the right function $g_\phi\in L^2$, and then I'll have the expression 
$$
\phi(f) = \langle f,g_\phi\rangle = \int_{\Bbb{R}^n} f(x)\bar{g}_\phi(x)dx
$$ 
The conjugate exponent expression 
$$
\frac{1}{p}+\frac{1}{q} = 1 
$$
comes from wanting to compute expressions that look just like $\langle f,g\rangle$: 
$$
\int_{\Bbb{R}^n} f(x) \bar{g}(x)dx
$$
but in the case where $f\in L^p$ for any $p$, not necessarily 2!  The reason we do this is that we also want to define bounded linear maps $\phi:L^p\rightarrow\Bbb{C}$, but if $f\in L^p$, the expression 
$$
\phi(f) = \int_{\Bbb{R}^n}f(x)\bar{g}(x)dx 
$$ will only define a bounded linear map (i.e. the product will be an $L^1$ integrable function for all $f\in L^p$) when $g\in L^{q}$ where $p$ and $q$ satisfy the conjugate exponent equation.  Another way of expressing this is that the dual of $L^p$ is $L^q$, where $1/p+1/q = 1$, i.e. a version of the Riesz representation theorem holds for $L^p$, but $g_\phi\in L^q$ instead of $L^p$.
See also Holder's inequality.
A: I think part of it is because of Holder's inequality, which dictates that
$$
|fg|_{1}\le |f|_{p}|g|_{q}
$$
But the other reasons goes into this is much deeper. For example, this duality is "structural" as it holds regardless of the measure space is $\sigma$-finite. And this has to do with concepts like uniform convexity. I used to wondering the same thing when I was learning the subject. For some related questions on this, see:
How should I prove the duality?
If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?
If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$
Is there a constructive proof of this characterization of $\ell^2$?
A: It's the only Hilbert space. That is, the norm comes from an inner product. Hilbert spaces have a lot of nice properties. For instance they are self-dual. They have a concept of orthogonality and angle. 
