[proofplan] We prove the Fredholm properties of $T = I - K$ in stages. First, compactness of $K$ forces $\ker(T)$ to be finite-dimensional (its unit ball is precompact). Second, we show $\operatorname{Range}(T)$ is closed by a compactness-based contradiction argument on the orthogonal complement of the kernel. Third, the closed-range property yields $\operatorname{Range}(T) = \ker(T^*)^\perp$ from the standard Hilbert space identity. Fourth, we prove the deepest fact -- injectivity implies surjectivity -- by constructing a strictly decreasing chain of iterated ranges and using Riesz's lemma to contradict compactness. Finally, we bootstrap the injectivity-implies-surjectivity result via a finite-rank perturbation to obtain $\dim \ker(T) = \dim \ker(T^*)$. [/proofplan] [step:Show that $\ker(T)$ is finite-dimensional via precompactness of the unit ball] [claim:Finite-dimensionality of the kernel] $\ker(T) = \ker(I - K)$ is a finite-dimensional subspace of $H$. [/claim] [proof] Let $B$ denote the closed unit ball of $\ker(T)$. For any $x \in B$, we have $Tx = 0$, equivalently $x = Kx$. Therefore $K(B) = B$: the identity map on $\ker(T)$ factors through $K$. Since $K$ is compact, $K(B)$ is precompact (its closure is compact in $H$). Thus the closed unit ball $B$ of the normed space $\ker(T)$ is compact. By the Riesz lemma characterisation of finite-dimensional spaces, a normed space whose closed unit ball is compact must be finite-dimensional. [/proof] [guided] We need to show $\ker(T)$ is finite-dimensional. The key mechanism is that compactness of $K$ constrains the "size" of the kernel by forcing the unit ball of $\ker(T)$ to be precompact. Start from the defining equation: if $x \in \ker(T)$, then $Tx = (I - K)x = 0$, which rearranges to $x = Kx$. This tells us that the identity map on $\ker(T)$ factors through the compact operator $K$. Let $B := \{x \in \ker(T) : \|x\| \le 1\}$ denote the closed unit ball of $\ker(T)$. For every $x \in B$, we have $x = Kx$, so $B = K(B)$. Now use the definition of compactness for operators: $K$ is compact means $K$ maps every bounded subset of $H$ to a precompact subset (one whose closure is compact). Since $B \subseteq \ker(T)$ is bounded ($\|x\| \le 1$ for all $x \in B$), the image $K(B)$ is precompact in $H$. But $K(B) = B$, so $B$ itself is precompact. Moreover, $B$ is closed (it is the closed unit ball of a normed subspace of $H$). A subset of a complete metric space that is both closed and precompact is compact. Therefore $B$ is compact. The Riesz lemma characterisation of finite-dimensional spaces states: a normed space is finite-dimensional if and only if its closed unit ball is compact. Since $B$, the closed unit ball of $\ker(T)$, is compact, we conclude $\dim \ker(T) < \infty$. Why does this argument fail without compactness of $K$? If $K$ is merely bounded, then $K(B) = B$ is bounded but not necessarily precompact. In an infinite-dimensional normed space, the closed unit ball is always bounded and closed, but never compact (this is precisely the content of the Riesz lemma in the other direction). So boundedness of $K$ alone cannot force $\ker(T)$ to be finite-dimensional. [/guided] [/step] [step:Prove $\operatorname{Range}(T)$ is closed by a normalisation-and-compactness argument] [claim:Closedness of the range] $\operatorname{Range}(T)$ is a closed subspace of $H$. [/claim] [proof] Let $(y_n)_{n \in \mathbb{N}} \subset \operatorname{Range}(T)$ with $y_n \to y$ in $H$. Write $y_n = Tx_n$. Using the orthogonal decomposition $H = \ker(T) \oplus \ker(T)^\perp$, write $x_n = u_n + v_n$ with $u_n \in \ker(T)$, $v_n \in \ker(T)^\perp$. Since $Tu_n = 0$, we have $Tv_n = y_n$. We claim $(v_n)$ is bounded. Suppose for contradiction that $\|v_n\| \to \infty$ (passing to a subsequence). Define the normalised vectors $w_n := v_n / \|v_n\|$. Then $\|w_n\| = 1$, each $w_n \in \ker(T)^\perp$, and: \begin{align*} \|Tw_n\| = \frac{\|Tv_n\|}{\|v_n\|} = \frac{\|y_n\|}{\|v_n\|} \to 0, \end{align*} since $(y_n)$ converges (hence is bounded) and $\|v_n\| \to \infty$. The relation $Tw_n = w_n - Kw_n \to 0$ gives $w_n - Kw_n \to 0$. Since $(w_n)$ is bounded and $K$ is compact, there exists a subsequence (still denoted $(w_n)$) with $Kw_n \to z$ for some $z \in H$. Then $w_n \to z$. By continuity of $T$: $Tz = \lim Tw_n = 0$, so $z \in \ker(T)$. But $\ker(T)^\perp$ is closed and $w_n \in \ker(T)^\perp$ for all $n$, so $z \in \ker(T)^\perp$. This gives $z \in \ker(T) \cap \ker(T)^\perp = \{0\}$, contradicting $\|z\| = \lim \|w_n\| = 1$. Therefore $(v_n)$ is bounded. Since $K$ is compact, passing to a subsequence gives $Kv_{n_j} \to w$ for some $w \in H$. From $v_{n_j} = Tv_{n_j} + Kv_{n_j} = y_{n_j} + Kv_{n_j}$, the subsequence $(v_{n_j})$ converges to $\bar{x} := y + w$. By continuity: $T\bar{x} = \lim Tv_{n_j} = \lim y_{n_j} = y$. So $y \in \operatorname{Range}(T)$. [/proof] [guided] This is the most technical step. We must show that $\operatorname{Range}(T)$ is closed, meaning every limit of elements in $\operatorname{Range}(T)$ is itself in $\operatorname{Range}(T)$. Let $(y_n) \subset \operatorname{Range}(T)$ converge to $y$. Write $y_n = Tx_n$. The challenge is that the preimages $x_n$ might be unbounded, making it impossible to extract a convergent subsequence. The solution is to project onto $\ker(T)^\perp$. Since $\ker(T)$ is a closed subspace (it is finite-dimensional by the previous step), we have the orthogonal decomposition $H = \ker(T) \oplus \ker(T)^\perp$. Write $x_n = u_n + v_n$ with $u_n \in \ker(T)$ and $v_n \in \ker(T)^\perp$. Since $Tu_n = 0$, we get $Tv_n = y_n$. The kernel component $u_n$ is irrelevant; we work with $v_n$. **Why are the $v_n$ bounded?** Suppose not: after passing to a subsequence, $\|v_n\| \to \infty$. Normalise: set $w_n = v_n / \|v_n\|$, so $\|w_n\| = 1$ and $w_n \in \ker(T)^\perp$. Then: \begin{align*} \|Tw_n\| = \frac{\|y_n\|}{\|v_n\|} \to 0. \end{align*} Writing $Tw_n = w_n - Kw_n \to 0$ means $w_n \approx Kw_n$ for large $n$. Since $\|w_n\| = 1$ and $K$ is compact, extract a subsequence with $Kw_n \to z$. Then $w_n \to z$ as well, and by continuity $Tz = 0$, so $z \in \ker(T)$. But $w_n \in \ker(T)^\perp$ and this subspace is closed, so $z \in \ker(T)^\perp$. Hence $z \in \ker(T) \cap \ker(T)^\perp = \{0\}$, contradicting $\|z\| = 1$. **Extracting the limit.** Now that $(v_n)$ is bounded, compactness of $K$ gives a subsequence with $Kv_{n_j} \to w$. Since $v_{n_j} = y_{n_j} + Kv_{n_j}$ (from $Tv_{n_j} = v_{n_j} - Kv_{n_j} = y_{n_j}$), the subsequence converges: $v_{n_j} \to y + w =: \bar{x}$. By continuity of $T$: \begin{align*} T\bar{x} = \lim_{j \to \infty} Tv_{n_j} = \lim_{j \to \infty} y_{n_j} = y. \end{align*} So $y \in \operatorname{Range}(T)$, and the range is closed. [/guided] [/step] [step:Characterise the range as $\operatorname{Range}(T) = \ker(T^*)^\perp$] [claim:Range equals orthogonal complement of adjoint kernel] $\operatorname{Range}(T) = \ker(T^*)^\perp$. [/claim] [proof] For any bounded linear operator on a [Hilbert space](/page/Hilbert%20Space), $\operatorname{Range}(T)^\perp = \ker(T^*)$: an element $y$ satisfies $(Tx, y)_H = 0$ for all $x \in H$ if and only if $(x, T^*y)_H = 0$ for all $x \in H$, which holds if and only if $T^*y = 0$. Taking orthogonal complements of both sides: $\overline{\operatorname{Range}(T)} = \ker(T^*)^\perp$. Since $\operatorname{Range}(T)$ is closed (previous step), $\overline{\operatorname{Range}(T)} = \operatorname{Range}(T)$, giving $\operatorname{Range}(T) = \ker(T^*)^\perp$. [/proof] [/step] [step:Prove injectivity implies surjectivity via iterated ranges and Riesz's lemma] [claim:Injectivity implies surjectivity] If $\ker(T) = \{0\}$, then $\operatorname{Range}(T) = H$. [/claim] [proof] Assume for contradiction that $\ker(T) = \{0\}$ but $\operatorname{Range}(T) \subsetneq H$. Define the iterated ranges: \begin{align*} H_0 := H, \qquad H_{n+1} := T(H_n) \quad \text{for } n \ge 0. \end{align*} Each $H_n$ is a closed subspace of $H$. This holds by induction: $H_0 = H$ is closed, and if $H_n$ is closed, then the restriction $T|_{H_n}: H_n \to H_{n+1}$ is an operator of the form $I - K|_{H_n}$ on a closed subspace, and the same compactness argument as in the previous closed-range proof (with $\ker(T|_{H_n}) = \{0\}$ by injectivity of $T$) shows $H_{n+1} = T(H_n)$ is closed. The chain is strictly decreasing: $H_1 = T(H) \subsetneq H_0$ by assumption. If $H_{n+1} = H_n$ for some $n \ge 1$, then $T$ maps $H_n$ onto $H_n$. Since $T$ is injective, $T|_{H_n}$ is bijective. By the same reasoning applied backwards through the chain, $T$ maps $H = H_0$ onto $H$, contradicting $H_1 \subsetneq H_0$. By Riesz's lemma, for each $n \ge 0$ there exists $u_n \in H_n$ with $\|u_n\| = 1$ and $\operatorname{dist}(u_n, H_{n+1}) \ge 1/2$. For $m < n$, writing $K = I - T$: \begin{align*} Ku_m - Ku_n = u_m - \underbrace{(Tu_m + u_n - Tu_n)}_{=:\, z}. \end{align*} We verify $z \in H_{m+1}$: since $u_m \in H_m$, we have $Tu_m \in T(H_m) = H_{m+1}$; since $n > m$, we have $u_n \in H_n \subseteq H_{m+1}$ and $Tu_n \in T(H_n) = H_{n+1} \subseteq H_{m+1}$; as $H_{m+1}$ is a subspace, $z \in H_{m+1}$. By the Riesz lemma property of $u_m$: \begin{align*} \|Ku_m - Ku_n\| = \|u_m - z\| \ge \operatorname{dist}(u_m, H_{m+1}) \ge \frac{1}{2}. \end{align*} The sequence $(Ku_n)$ has no convergent subsequence (all pairwise distances are $\ge 1/2$), yet $(u_n)$ is bounded ($\|u_n\| = 1$). This contradicts compactness of $K$. [/proof] [guided] This is the heart of the Fredholm alternative. The claim is that for $T = I - K$ with $K$ compact, injectivity of $T$ forces surjectivity -- a property that fails for general bounded operators on infinite-dimensional spaces (consider the right shift on $\ell^2$). **Building the chain.** Assume $\ker(T) = \{0\}$ but $T$ is not surjective. Define iterated images: \begin{align*} H_0 := H, \qquad H_{n+1} := T(H_n). \end{align*} Each $H_n$ is closed (by the same compactness argument used to show $\operatorname{Range}(T)$ is closed, applied to $T$ restricted to $H_n$; the injectivity of $T$ means $\ker(T|_{H_n}) = \{0\}$). **Why is the chain strictly decreasing?** We have $H_1 \subsetneq H_0$ by the non-surjectivity assumption. If the chain stabilised at some level, say $H_{n+1} = H_n$, then $T$ would map $H_n$ bijectively onto itself. Propagating backwards through the chain, this would force $T$ to map $H_0 = H$ onto itself, contradicting $H_1 \subsetneq H_0$. **The Riesz lemma construction.** Since $H_{n+1} \subsetneq H_n$ and both are closed, Riesz's lemma provides unit vectors $u_n \in H_n$ with $\operatorname{dist}(u_n, H_{n+1}) \ge 1/2$ for each $n$. **Contradicting compactness.** For $m < n$, we compute $Ku_m - Ku_n = (I - T)u_m - (I - T)u_n = u_m - (Tu_m + u_n - Tu_n)$. Define $z := Tu_m + u_n - Tu_n$. We check that $z \in H_{m+1}$: - $Tu_m \in T(H_m) = H_{m+1}$, - $u_n \in H_n \subseteq H_{m+1}$ (since $n > m$), - $Tu_n \in T(H_n) = H_{n+1} \subseteq H_{m+1}$. Since $H_{m+1}$ is a linear subspace, $z \in H_{m+1}$. Therefore: \begin{align*} \|Ku_m - Ku_n\| = \|u_m - z\| \ge \operatorname{dist}(u_m, H_{m+1}) \ge \frac{1}{2}. \end{align*} So every pair of terms in the sequence $(Ku_n)$ is at least $1/2$ apart. No subsequence can be Cauchy, hence no subsequence converges. But $(u_n)$ is bounded ($\|u_n\| = 1$ for all $n$) and $K$ is compact, so $(Ku_n)$ must have a convergent subsequence. This is a contradiction. [/guided] [/step] [step:Establish the dimension equality $\dim \ker(T) = \dim \ker(T^*)$ via finite-rank perturbation] [claim:Dimension equality] $\dim \ker(T) = \dim \ker(T^*)$. [/claim] [proof] Since $K^*$ is also compact, all previous results apply equally to $T^* = I - K^*$. We prove $d := \dim \ker(T) \le d^* := \dim \ker(T^*)$; the reverse inequality follows by applying the same argument to $T^*$ (using $(T^*)^* = T$). Suppose for contradiction that $d > d^*$. Choose an orthonormal basis $\{e_1, \ldots, e_d\}$ for $\ker(T)$ and an orthonormal basis $\{f_1, \ldots, f_{d^*}\}$ for $\ker(T^*)$. Define a finite-rank operator: \begin{align*} F: H &\to H \\ x &\mapsto \sum_{j=1}^{d^*} (x, e_j)_H \, f_j. \end{align*} Set $\tilde{T} := T + F = I - (K - F)$. Since $F$ is finite-rank, $K - F$ is compact, so all previous results apply to $\tilde{T}$. **$\tilde{T}$ is surjective:** Let $y \in H$. Decompose $y = y_1 + y_2$ with $y_1 \in \ker(T^*)^\perp = \operatorname{Range}(T)$ and $y_2 \in \ker(T^*)$. Choose $x_1 \in \ker(T)^\perp$ with $Tx_1 = y_1$; then $Fx_1 = 0$ since $(x_1, e_j)_H = 0$ for all $j$. Write $y_2 = \sum_{j=1}^{d^*} \beta_j f_j$ and set $e := \sum_{j=1}^{d^*} \beta_j e_j \in \ker(T)$. Then $Te = 0$ and $Fe = y_2$, so: \begin{align*} \tilde{T}(x_1 + e) = Tx_1 + Fx_1 + Te + Fe = y_1 + 0 + 0 + y_2 = y. \end{align*} **$\tilde{T}$ is not injective:** Since $d > d^*$, the vector $e_{d^*+1} \in \ker(T)$ satisfies $Te_{d^*+1} = 0$ and $Fe_{d^*+1} = \sum_{j=1}^{d^*} (e_{d^*+1}, e_j)_H f_j = 0$ by orthonormality. So $\tilde{T}e_{d^*+1} = 0$ with $e_{d^*+1} \ne 0$. **Contradiction:** Surjectivity of $\tilde{T}$ means $\operatorname{Range}(\tilde{T}) = H$, so $\ker(\tilde{T}^*) = \{0\}$ (from the range characterisation). By the injectivity-implies-surjectivity result applied to $\tilde{T}^* = I - (K^* - F^*)$: $\ker(\tilde{T}^*) = \{0\}$ implies $\operatorname{Range}(\tilde{T}^*) = H$. Then $\ker(\tilde{T})^\perp = \operatorname{Range}(\tilde{T}^*) = H$, giving $\ker(\tilde{T}) = \{0\}$. But $e_{d^*+1} \in \ker(\tilde{T}) \setminus \{0\}$ -- a contradiction. [/proof] [guided] This is the bootstrapping step: we use the injectivity-implies-surjectivity result to prove that the kernel dimensions on both sides match. Since $K^*$ is compact whenever $K$ is, all the Fredholm machinery applies to $T^* = I - K^*$ as well. It suffices to show $d := \dim \ker(T) \le d^* := \dim \ker(T^*)$; applying the same argument with $T$ and $T^*$ swapped (using $(T^*)^* = T$) gives the reverse inequality. Suppose $d > d^*$. We will construct an operator $\tilde{T} = I - \tilde{K}$ (with $\tilde{K}$ compact) that is surjective but not injective, and derive a contradiction. **The perturbation.** Pick orthonormal bases $\{e_1, \ldots, e_d\}$ for $\ker(T)$ and $\{f_1, \ldots, f_{d^*}\}$ for $\ker(T^*)$. Define: \begin{align*} F: H &\to H, \\ x &\mapsto \sum_{j=1}^{d^*} (x, e_j)_H \, f_j. \end{align*} This is a finite-rank operator (its range is $\operatorname{span}\{f_1, \ldots, f_{d^*}\}$, which has dimension $d^*$). Set $\tilde{T} := T + F = I - (K - F)$. Since $F$ is finite-rank, it is compact, so $K - F$ is compact and all Fredholm results apply to $\tilde{T}$. **Surjectivity of $\tilde{T}$.** The idea is that $F$ "fills in the gaps" left by $T$. Any $y \in H$ decomposes as $y = y_1 + y_2$ with $y_1 \in \operatorname{Range}(T) = \ker(T^*)^\perp$ and $y_2 \in \ker(T^*)$. For $y_1$, choose a preimage $x_1 \in \ker(T)^\perp$ (so $Tx_1 = y_1$ and $Fx_1 = 0$). For $y_2 = \sum \beta_j f_j$, set $e = \sum \beta_j e_j \in \ker(T)$; then $Te = 0$ and $Fe = y_2$. So $\tilde{T}(x_1 + e) = y$. **Non-injectivity of $\tilde{T}$.** Since $d > d^*$, the basis vector $e_{d^*+1}$ lies in $\ker(T)$ but is orthogonal to $e_1, \ldots, e_{d^*}$, so $Fe_{d^*+1} = 0$ and $\tilde{T}e_{d^*+1} = 0$. **The contradiction chain.** Surjectivity gives $\operatorname{Range}(\tilde{T}) = H$, so $\ker(\tilde{T}^*) = \{0\}$. By injectivity-implies-surjectivity for $\tilde{T}^*$: $\operatorname{Range}(\tilde{T}^*) = H$. Then $\ker(\tilde{T})^\perp = \operatorname{Range}(\tilde{T}^*) = H$, forcing $\ker(\tilde{T}) = \{0\}$. But $e_{d^*+1} \in \ker(\tilde{T})$ is nonzero -- contradiction. [/guided] [/step] [step:Assemble the four Fredholm properties and derive the alternative] Combining all results: $\ker(T)$ is finite-dimensional (Step 1), $\operatorname{Range}(T)$ is closed (Step 2), $\operatorname{Range}(T) = \ker(T^*)^\perp$ (Step 3), and $\dim \ker(T) = \dim \ker(T^*)$ (Step 5). If $\ker(T) = \{0\}$, then $\dim \ker(T^*) = 0$ by the dimension equality, so $\ker(T^*) = \{0\}$ and $\operatorname{Range}(T) = \{0\}^\perp = H$. Thus $T$ is bijective. Since $T$ is a bounded bijection between Banach spaces, the [Open Mapping Theorem](/theorems/631) provides a bounded inverse, and $Tx = y$ has a unique solution for every $y \in H$. If $\ker(T) \ne \{0\}$, then $Tx = y$ is solvable if and only if $y \in \operatorname{Range}(T) = \ker(T^*)^\perp$, i.e., $(y, v)_H = 0$ for all $v \in \ker(T^*)$. [/step]