Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open and $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Omega)}\;\;\;\text{for }u,v\in H_0^1(\Omega).$$ By the Lax-Milgram theorem, $$\forall f\in L^2(\Omega):\exists! v_f\in H_0^1(\Omega):\forall u\in H_0^1(\Omega):\mathfrak a(u,v_f)=\langle u,f\rangle_{L^2(\Omega)}\tag1.$$ The map $$S:L^2(\Omega)\to H_0^1(\Omega)\;,\;\;\;f\mapsto v_f$$ is
- a bounded linear operator from $L^2(\Omega)$ to $H_0^1(\Omega)$; and
- a compact self-adjoint linear operator on $L^2(\Omega)$.
By (2.) and the spectral theorem, we know that there is an orthonormal basis $(e_i)_{i\in I}\subseteq H_0^1(\Omega)$ of the closed subspace $\overline{\mathcal R(S)}$ of $L^2(\Omega)$, where $I:=\mathbb N\cap[0,\operatorname{rank}S]$, such that $$S=\sum_{i\in I}\mu_ie_i\otimes e_i,\tag2$$ where the sum converges in $\mathfrak L(L^2(\Omega))$, for some nondecreasing $(\mu_i)_{i\in I}$.
As usual, $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $L^2(\Omega)$ by an orthonormal basis $(e_n)_{n\in\mathbb N\setminus I}$ of $\mathcal N(S)$ and we may set $\mu_n:=0$ for $n\in\mathbb N\setminus I$.
Question 1: Is it actually possible that $I\ne\mathbb N$?
Now let $$P:=-\Delta:C_c^\infty(\Omega)\to C_c^\infty(\Omega)$$ and set $\lambda_n:=\frac1{\mu_n}$ for $n\in\mathbb N$ with the convention that $\frac10:=\infty$. We easily see that it holds $$Pe_i=\lambda_i\varphi\tag3$$ in the weak sense, i.e. $$\langle e_i,P\varphi\rangle_{L^2(\Omega)}=\lambda_i\langle e_i,\varphi\rangle_{L^2(\Omega)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\tag4,$$ for all $i\in I$ (even for all $i\in\mathbb N\setminus I$?).
Now, from Riesz' representation theorem, we see that there is a unique bounded linear operator $A$ on $H_0^1(\Omega)$ such that $$\mathfrak a(\;\cdot\;,v)=\langle\;\cdot\;,Av\rangle_{H^1(\Omega)}\;\;\;\text{for all }v\in H_0^1(\Omega)\tag5$$ and it is easy to see that $A$ is self-adjoint.
Question 2: Are we able to show that $\lambda_i$ is an eigenvalue of $A$ for all $i\in I$? And if we actually can, does $A$ admit any eigenvalue not contained in $\{0\}\cup\{\lambda_i:i\in I\}$?$^1$
From $(4)$ and the definition of $A$ we see that $$\langle\lambda_ie_i,\varphi\rangle_{L^2(\Omega)}=\langle Ae_i,\varphi\rangle_{H^1(\Omega)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\tag6,$$ but that doesn't allow us to conclude, since different inner products are taken on the left- and right-hand side.
So, maybe we need to consider the linear operator $$B:=\mathfrak a(u,\;\cdot\;)\;\;\;\text{for }u\in H_0^1(\Omega)$$ instead?
In any case, what I finally would like to understand is why it makes sense to say that $$e^{t\Delta}f:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle f,e_n\rangle_{L^2(\Omega)}e_n\;\;\;\text{for }f\in L^2(\Omega)\tag7$$ is "the semigroup generated by the Dirichlet Laplacian on $\Omega$.
Please note that I'm familiar with semigroups and their generators. But I actually only know the definition of the operator exponential $e^{tT}$ for a bounded linear operator $T$. I guess the notation of the left-hand side in $(7)$ has to be understood with the "functional calculus" sense used in the spectral theorem for bounded linear self-adjoint operators ... But which bounded linear self-adjoint operator, corresponding to the Laplacian, are we considering here (if it's not $A$)?
$^1$ $A$ is certainly not compact and hence its spectrum $\sigma(A)$ should not be equal to the point spectrum $\sigma_p(A)$ (unless I'm terribly wrong). But that doesn't exclude the possibility that $\sigma_p(A)$ is countable ... So, maybe we can actually show that $\sigma_p(A)\setminus\{0\}=\{\lambda_i:i\in I\}$.
Q1: the range of $S$ is $H^1_0$, which is dense in $L^2$, so the sum (2) has to be an infinite sum.
(3) makes no sense as $e_i\not\in C_c^\infty$ in general
Q2: The difference between (5) and the eigenvalues of $S$ is: (5) means $$ \int_\Omega \nabla u\nabla v = \lambda \int_\Omega \nabla u\nabla v + uv \quad \forall v\in H^1_0(\Omega) $$ while the eigenvalues of $S$ are defined by $$ \int_\Omega \nabla u\nabla v = \lambda \int_\Omega uv\quad \forall v\in H^1_0(\Omega). $$ The difference is in the inner products to form the right-hand side. $S$ is compact, but $A:H^1_0\to H^1_0$ is bijective.