Dimension of a certain $L^p$ quotient space. Define $L^p_0 := \{ f \in L^p : \int f = 0 \}$. I am trying to calculate the dimension of the cokernel of the inclusion operator $i:L^p_0 \to L^p$. That is, I am trying to calculate $$\dim ( L^p / L^p_0 )$$
If $p=2$, I would have a Hilbert space and I would say that 
$$
L^2_0 = \{ f \in L^2 : \langle f,c\rangle =0\},
$$ 
that is, $L^2_0$ is the orthogonal complement of the subspace of constant functions. As this subspace has dimension 1, I guess there is some simple argument to conclude that the cokernel of its orthogonal space is also 1 (Is this because of the projection theorem? How can I do this rigorously?)
For the general case, I don't know how to proceed to do the same thing rigorously for a Banach space. I think this is pretty straightforward but I've forgotten most of functional analysis stuff and I'd like to re-learn these things. Thanks in advance.
Bonus question: Is it true that the dual space $(L^p_0)^* = L^{p'} / \mathbb{R}$ ? Is the inclusion $i:L^p_0 \to L^p$ the adjoint operator of the projection $pr: L^{p'} \to L^{p'} / \mathbb{R}$ ?
 A: Generally, if $T:X\to Y$ is a surjective linear map, then $X/\ker T$ is isomorphic to $Y$. Applied to your case, $X=L^p$, $Y=\mathbb{R}$, $T=\int f$, this yields $L^p/L^p_0 $ being isomorphic to $\mathbb{R}$.   (Or $\mathbb{C}$ if you use complex scalars)
Bonus content:


*

*If $M$ is a closed subspace of Banach space $X$, the inclusion $i:M\to X$ has adjoint $i^*:X^*\to M^*$, which is simply the restriction of linear functionals on $X$ to $M$. This adjoint is surjective by the Hahn–Banach theorem. Its kernel is $M^\perp\subset X^*$, the space of functionals  vanishing on $M$. It follows that $M^* $ is (isometrically) isomorphic to $X^*/M^\perp$.

*Similarly (and more relevant to your question), the projection map $\pi : X\to X/M$ has adjoint $\pi^*:(X/M)^*\to X^*$ which acts by $\pi^*(f)(x) = f(x+M)$. The range of $\pi$ is precisely $M^\perp$. Indeed, it's clear that $\pi^*(f)$ vanishes on $M$; and conversely, every element $f\in M^\perp$ naturally defines a linear functional on the coset space $X/M$. Lastly, $\pi^*$ is an isometry, which one can see using the definition of the quotient norm on $X/M$. 
In your example, $M=L_0^p$, and $M^\perp = \{f\in L^{p'} : f\equiv \operatorname{const}\}$. So, $(X/M)^*$ is isomorphic to the space of constant functions in $L^{p'}$, which is isomorphic to $\mathbb{R}$.
Suggested reading: Conway, A first course in functional analysis, the chapter on Banach spaces.
