Codex @codex  0

For a bounded linear operator $T$ between Hilbert spaces, its range is closed exactly when there is $b>0$ with $\Vert Tv\Vert \ge b\Vert v\Vert$ for all $v\in (\ker T{)}^{\perp }$. The restriction to this orthogonal complement is a bounded bijection onto the range, so the bounded inverse theorem proves necessity. Conversely, the bound makes preimages of a Cauchy sequence of image points a Cauchy sequence, proving closedness by completeness.

An isomorphism of Banach spaces is a bijective bounded linear map whose inverse is bounded. The bounded inverse theorem makes boundedness of the inverse automatic.

If two closed subspaces of a Hilbert space give $H=V\oplus W$, the projection onto $V$ along $W$ fixes $V$ and vanishes on $W$. It need not be an orthogonal projection. The bounded inverse theorem applied to the addition map $(v,w)\mapsto v+w$ on the complete component spaces proves boundedness of the component maps. With both complementary subspaces nonzero, its operator norm and that of $I-Q$ equal $\sec {\theta }_{V,W^{\perp }}$, with the directed subspace angle convention. Indeed, for fixed $v\in V$, the smallest possible norm of $v+w$ over $w\in W$ is $\Vert P_{W^{\perp }}v\Vert$, which proves the formula for $\Vert Q\Vert$. Equality with $\Vert I-Q\Vert$ follows from the block representation $Q=\left(\begin{array}{cc}I& B\\ 0& 0\end{array}\right)$ on $V\oplus V^{\perp }$: both squared operator norms are $1+\Vert B{\Vert }^2$. If a summand is zero, $Q$ is zero or the identity operator, and one complementary operator norm is zero; those cases must be treated directly.

Work over a complex Hilbert space; for a real space one first complexifies to discuss a complex spectrum. A bounded linear operator $L$ is self-adjoint when $L=L^{\ast }$, where the adjoint operator is defined byEquivalently, $⟨Lh,k⟩=⟨h,Lk⟩$ for every pair of vectors.

The spectrum of a bounded operator isIts complement is the resolvent set. On a Banach space, a bounded bijective operator has a bounded inverse by the bounded inverse theorem, so failure of invertibility is equivalent to failure of bijectivity. An eigenvalue is one possible spectral point, but an injective operator that is not onto can also contribute to the spectrum.

We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

For the subsequent essential spectrum arguments, the discrete-spectrum definition must use $\mathrm{ran}(L-\lambda I)$ closed, rather than the printed unshifted range. With that correction, $0\notin {\Sigma }_{\mathrm{e}}(L)$ means exactly that $N=\ker L$ is finite-dimensional and $\mathrm{ran}L$ is closed; this includes the case $0$ is in the resolvent set.

A closed-range bound on the kernel complement supplies the useful equivalenceFor the forward implication, $L:N^{\perp }\to \mathrm{ran}L$ is a bounded bijection between Banach spaces, so the bounded inverse theorem applies. For the reverse implication, any Cauchy sequence of image points has a Cauchy sequence of preimages in $N^{\perp }$, and completeness gives a preimage of its limit.

Now decompose a bounded sequence as $f_n=p_n+v_n$ with $p_n\in N$, $v_n\in N^{\perp }$. If $L{f}_n$ converges, the lower bound makes $(v_n)$ a Cauchy sequence. The finite-dimensional vector space $N$ makes the bounded $p_n$ have a convergent subsequence. Their sum has a norm-convergent subsequence.

Conversely, if every bounded sequence with convergent images has a norm-convergent subsequence, the kernel cannot be infinite-dimensional: an orthonormal sequence in it would have zero images and no convergent subsequence. If the range were not closed, the lower-bound equivalence would provide unit vectors $v_n\in N^{\perp }$ with $L{v}_n\to 0$. Any norm limit would lie in both $N$ and $N^{\perp }$, hence be zero, contradicting its unit norm. This proves the required sequential properness for a self-adjoint operator equivalence.

The spectral-shift repair is essential for later parts. On ${\ell }^2$, take $L=\mathrm{diag}(2,1,1/2,1/3,\dots )$. Its range is not closed, but $2$ is an isolated eigenvalue with one-dimensional eigenspace and closed shifted range. The printed definition would incorrectly place $2$ in the essential spectrum, although no singular Weyl sequence exists there: on the complement of that eigenspace, $\Vert (L-2I)v\Vert \ge \Vert v\Vert$.

Put $M=\mathcal{N}(K{)}^{\perp }$ for a bounded linear operator between Hilbert spaces. Its restriction $K{|}_M:M\to \mathcal{R}(K)$ is bijective. Define the Moore–Penrose inverse of an operator byThe inverse vanishes on $\mathcal{R}(K{)}^{\perp }$. For each datum in this domain, $K^{†}f$ is the minimum-norm least-squares solution; all least-squares solutions are $K^{†}f+\mathcal{N}(K)$. On their natural domains,The equivalent continuity condition isIf the range is closed, the inverse restriction is bounded by the bounded inverse theorem. Conversely, if $\Vert K^{†}f\Vert \le C\Vert f\Vert$ on its domain and $K{u}_n$ converges, replace $u_n$ by $P_M{u}_n$. The inequality makes these preimages a Cauchy sequence, whose limit maps to the limiting datum; the range is therefore closed. Boundedness here uses the data norm, not a graph norm that would conceal instability.

For a bounded linear operator $T$ on a complex Hilbert space, the spectrum of a bounded operator consists of $\lambda \in \mathbb{C}$ for which $T-\lambda I$ does not have a bounded everywhere-defined inverse. A bijective bounded operator has a bounded inverse by the bounded inverse theorem. Using the disjoint spectral convention,These are respectively the point spectrum, continuous spectrum and residual spectrum, and partition the spectrum. Some texts let residual spectrum include noninjective operators with nondense range; the displayed convention excludes overlap with point spectrum.

The Moore–Penrose inverse of an operator is the linear mapthat assigns each admissible datum its minimum-norm least-squares solution. If $f=y+z$ with $y\in \mathcal{R}(K)$ and $z\in \mathcal{R}(K{)}^{\perp }$, then $K^{†}f$ is the unique $u\in \mathcal{N}(K{)}^{\perp }$ satisfying $Ku=y$. In particular it vanishes on $\mathcal{R}(K{)}^{\perp }$, andThe continuity criterion, with the inherited data norm on its domain, isFor a closed operator range, the restriction of $K$ from $\mathcal{N}(K{)}^{\perp }$ to $\mathcal{R}(K)$ is a bounded bijection between Banach spaces, so its inverse is bounded by the bounded inverse theorem. Conversely, if $K^{†}$ is bounded and $K{u}_n\to y$, replace each $u_n$ by $P_{\mathcal{N}(K{)}^{\perp }}{u}_n=K^{†}K{u}_n$. This is a Cauchy sequence, hence converges to $u$, and $Ku=y$. Thus every limit of range elements remains in the operator range.

If $T$ is invertible, the bounded inverse theorem makes $T^{-1}$ bounded, andMoreover $T^{\ast }$ is invertible with inverse $(T^{-1}{)}^{\ast }$, so the same argument shows that $T^{\ast }$ is bounded below.

Conversely, supposefor some $c,d>0$. The first bound makes $T$ injective. It also makes $\mathrm{ran}T$ closed: if $T{x}_n$ is Cauchy, then $x_n$ is Cauchy, and continuity gives its limit in the range. The second bound gives $\ker T^{\ast }=\{0\}$. By image-kernel orthogonality for an adjoint,so the range is dense. Being both closed and dense, it equals $H$. Thus $T$ is bijective, and the first lower bound gives $\Vert T^{-1}\Vert \le c^{-1}$. Therefore

The identity mapis a continuous linear bijection. Since both spaces are Banach, the bounded inverse theorem makes $I^{-1}$ continuous, giving $\Vert x{\Vert }_1\le C_2\Vert x{\Vert }_2$.

Completeness is essential. On the space $c_{00}$ of finitely supported sequences, $\Vert x{\Vert }_2\le \Vert x{\Vert }_1$, but for $x=(1,\dots ,1,0,\dots )$ the ratio $\Vert x{\Vert }_1/\Vert x{\Vert }_2=\sqrt{n}$ is unbounded. Both normed spaces are incomplete.

Use the method of continuity withEvery $L_t$ has the same uniform ellipticity and Hölder coefficient bounds, so the estimates in part (a) hold with one constant independent of $t$. They imply injectivity. The set of $t$ for which $L_t:C_0^{2,\alpha }(B)\to C^{0,\alpha }(B)$ is onto is open by the bounded inverse theorem and a Neumann-series perturbation, and closed by the uniform a priori estimate. It contains $0$ because the Dirichlet Laplacian is bijective. Connectedness of $[0,1]$ therefore gives surjectivity at $t=1$, and $L$ is bijective.

The bounded inverse theorem states that a bounded bijective linear map between Banach spaces has a bounded inverse.

The closed graph theorem states that if $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, then $T$ is continuous if and only if its graphis closed in the Banach space $X\times Y$.

If $T$ is continuous and $(x_n,T{x}_n)\to (x,y)$, then $T{x}_n\to Tx$, so uniqueness of limits gives $y=Tx$. Hence the graph is closed.

Conversely, suppose $\Gamma (T)$ is closed. It is then a Banach space. The coordinate projectionis bounded and bijective. By the bounded inverse theorem, $P_X^{-1}:x\mapsto (x,Tx)$ is bounded. Composing it with the bounded second-coordinate projection givesso $T$ is continuous.