Step 1: Dichotomy

Suppose $(I - \lambda K)$ is not injective. Then $\ker(I - \lambda K) \neq \{0\}$. We show that alternative (2) holds.

Conversely, if $(I - \lambda K)$ is injective, we must show it is surjective (alternative 1).

Step 2: Kernel is Finite-Dimensional

Assume $\dim \ker(I - \lambda K) = \infty$. Choose an orthonormal sequence $(e_n)$ in $\ker(I - \lambda K)$. Then $(I - \lambda K)e_n = 0 \implies e_n = \lambda K(e_n)$

Since $K$ is compact, $(K(e_n))$ has a convergent subsequence. But $K(e_n) = e_n/\lambda$, and $(e_n)$ is orthonormal with $\|e_n - e_m\|^2 = 2$ for $n \neq m$, a contradiction.

Therefore $\dim \ker(I - \lambda K) < \infty$.

Step 3: Range is Closed

Let $y_n = (I - \lambda K)x_n$ with $y_n \to y$. We show $y$ is in the range.

Decompose $x_n = v_n + w_n$ where $v_n \in \ker(I - \lambda K)$ and $w_n \in (\ker(I - \lambda K))^\perp$.

From $(I - \lambda K)x_n = y_n$, we have $(I - \lambda K)w_n = y_n$ (since $v_n$ is in the kernel).

By a compactness argument using the restriction of $(I - \lambda K)$ to the finite-codimensional space $(\ker(I - \lambda K))^\perp$, we can show $(w_n)$ is bounded, hence has a subsequence with $K(w_{n_k}) \to z$.

Then $w_{n_k} = y_{n_k} + \lambda K(w_{n_k}) \to y + \lambda z$, so $y = (I - \lambda K)(y + \lambda z)$.

Step 4: Injectivity Implies Surjectivity

Assume $(I - \lambda K)$ is injective. Since the range is closed and has finite codimension (by Fredholm theory), and the kernel is trivial, we have codimension zero, so the range is all of $X$.

Step 5: Adjoint Relationship

The dimension formula $\dim \ker(I - \lambda K) = \dim \ker(I - \lambda K^*)$ follows from general Fredholm theory: for operators of the form $I - K$ with $K$ compact, the index is zero.