I am currently trying to understand a proof of the following Claim
Let $(\mathcal{M},g)$ be a compact, oriented Riemannian manifold and let $f:\mathcal{M} \to \mathbb{R}$ be a function. Then $$ \Delta_g u = f $$ is solvable if and only if $f$ has zero mean. (By which I mean that $\int f d \mu_g = 0 $, where $d \mu_g$ denotes the volume form on the manifold)
To prove the 'if'-direction the proof introduces the energy functional $$ E[u]:= \int_{\mathcal{M}} \left\{ \frac12 \left|du\right|^2_g+fu \right\} d \mu_g, $$ where $\left|du\right|^2_g:= g^{ij}\partial_iu \partial_j u$ and one can then see that a minimum of $E$ corresponds to a solution of the aforementioned PDE using the Euler-Lagrange equations. Now the proof continues by showing that $E$ is bounded from below and then immediately jumps to the conclusion that $E$ must have a minimum in $H_1(\mathcal{M})$.
My first question concerns the last step. How does this follow? The bound from below is given by $$ E[u] \geq - \frac{1}{2 \lambda_1} \int_{\mathcal{M}} f^2 d \mu_g, $$ if this is of any moment.
My second question is connected to uniqueness for this PDE. I heard that the solution $u$ will be unique up to an additive constant and tried to prove this in the standard way by assuming I find two solutions $u, \tilde u$ which then would satisfy $ \Delta_g (u-\tilde u) = 0$ and thus I would have $$ \int_{\mathcal{M}} \frac12 \left|d(u-\tilde u)\right|^2_g d \mu_g = 0. $$ By the positivity of the expression in the integral $ \left|d(u-\tilde u)\right|^2_g = 0$. But here I am stuck. How to proceed?
In my eyes, it would be easier to use the Lemma of Lax-Milgram to study exitence and uniqueness of your PDE, but anyway.
Since your functional is bounded from below, it has an $\inf$. Now, because it is defined on a reflexive space, this $\inf$ is in fact a $\min$.
Concerning your second question: you get more than $|d(u-\tilde{u})|^2_g=0$ pointwise: in fact, you do get that the $L^2(\mathcal M)$-norm of $d(u-\tilde{u})$ vanishes, hence $d(u-\tilde{u})=0$, i.e., $u-\tilde{u}$ is a constant.