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For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.

The Picard criterion with the input/output convention above statesNecessity follows from $⟨f,v_j⟩={\sigma }_j⟨u,u_j⟩$ and Bessel inequality. Conversely the summability constructs $u={\sum }_j{\sigma }_j^{-1}⟨f,v_j⟩u_j\in U$; the closure condition ensures that its image is all of $f$. The closure condition must not be omitted: a component orthogonal to the range cannot be reconstructed.

For the diagonal example, $f_j=j^{-p}$ is itself in ${\ell }^2$ only for $p>1/2$. Its preimage would be $u_j=j^{1-p}$, so the range criterion is the convergence of ${\sum }_j{j}^{2-2p}$. The P-series givesAt $p=3/2$ the reconstruction norm diverges harmonically. For $1/2<p\le 3/2$ the data are legitimate but have no preimage; for $0<p\le 1/2$ they are not even in the specified data space.

The range of the Volterra integration operator is $\{g\in H^1(0,1):g(0)=0\}$, using the representative of a Sobolev space element that is an absolutely continuous function. Indeed $Ku$ has weak derivative $u$ and zero initial trace; conversely the fundamental theorem of calculus reconstructs such a $g$ from its derivative. This range contains smooth compactly supported functions and is dense in $L^2(0,1)$.

The given step is in $L^2$, and its value at the single midpoint is immaterial. It cannot be the image of an $L^2$ function: such an image is continuous, whereas no continuous representative agrees almost everywhere with zero on the left half and one on the right half. Its distributional derivative is a Dirac delta distribution, not an $L^2$ function. ThusFor a direct approximation, replace the jump by a linear ramp of width $1/n$ centered at $1/2$. Each ramp starts at zero and has an $L^2$ derivative, so lies in the range, while its squared $L^2$ error is $1/(12n)$. Its derivative norm is $\sqrt{n}$, illustrating unstable differentiation.

The supplied SVD gives $K({\sigma }_j{u}_j)={\sigma }_j^2{v}_j$, or more simply $K^{†}({\sigma }_j{v}_j)=u_j$. Since ${\sigma }_j{v}_j\to 0$ but $\Vert u_j\Vert =1$, the Moore–Penrose inverse is discontinuous. Its domain is the dense, nonclosed range above, and there it is the weak derivative.

In an infinite-dimensional Hilbert space, every bounded sequence has a weakly convergent subsequence, but it need not have a norm-convergent one. For example an orthonormal sequence $e_n$ has pairwise distance $\sqrt{2}$ and therefore no strongly convergent subsequence, although $e_n\rightharpoonup 0$ by Bessel inequality.

More generally, weak sequential compactness of bounded sequences in a reflexive Banach space ensures a weakly convergent subsequence with its limit in that space. Conversely, this property for all bounded sequences characterizes reflexivity, using the Eberlein-Šmulian theorem and weak compactness of the closed unit ball. Thus reflexivity replaces finite-dimensional norm compactness by weak sequential compactness.

In a nonreflexive Banach space there is no such general weak-subsequence guarantee. The coordinate unit vectors in ${\ell }^1$ give an example: by the Schur property, weak convergence would imply norm convergence, impossible for their pairwise distance two. Banach-Alaoglu theorem supplies weak-star compactness of a dual unit ball, which gives convergent subnets, not automatically convergent subsequences. For a separable predual, the weak-star compact dual ball is metrizable and subsequences do suffice. For an arbitrary Banach space one can also regard a bounded sequence in its bidual and extract a weak-star convergent subnet there; the limit need not belong to the original space. These qualifications distinguish weak from weak-star compactness.

First use the usual Hilbert space convention for the proximal operator, identifying the dual element $z$ with its representing vector. For $c\ne 0$, completing the square givesSet $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\Vert v-[c(q-z)-y]{\Vert }^2/2+\alpha c^{2}J(v)$. Thus the proximal operator under affine rescaling isNegative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.

The PDF actually assumes an arbitrary Banach space, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define ${\mathrm{Prox}}_F(s)=\arg {\min }_v\{F(v)+\Vert v-s{\Vert }^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the norm still gives the valid change of variablesThis retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$:For a concrete failure of the Hilbert shortcut, take $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\le 0$ with $x_1+x_2=-1$ minimizes $\Vert x{\Vert }_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique. For $c=0$ with finite constant penalty the Banach condition is $-z\in \mathcal{J}(x-q)$, not generally $x=q-z$.

Because $w_k\ge c>0$, a finite sum ${\sum }_k{w}_k|u_k|$ implies $u\in {\ell }^1$. Thus the printed weighted one-norm on l2 equals the extended nonnegative sum on every $u\in {\ell }^2$. Unbounded weights may make that sum infinite for some $u\in {\ell }^1$; membership in ${\ell }^1$ alone is not a finiteness assertion.

For each $N$, $J_N(u)={\sum }_{k=1}^N{w}_k|u_k|$ is continuous in the l2 sequence space, and $J={\sup }_N{J}_N$. If $u^{(n)}\to u$ in norm, then for every $N$,Taking the supremum over $N$ provesIt is also weakly lower semicontinuous by the same finite-coordinate argument, and convex and proper because zero and finite sequences have finite penalty.

At a finite-penalty base point $v$, coordinate variations give the subdifferential of a weighted one-norm on l2:These conditions are sufficient by summing the coordinate supporting inequalities. Since $w_k\ge c$, an infinitely supported $v$ would require infinitely many $|p_k|\ge c$, contradicting $p\in {\ell }^2$. Thus a finite-penalty point can have empty subdifferential; a Bregman base with an actual subgradient must be finitely supported.

For any such $p$, $⟨p,v⟩=J(v)$ and the Bregman distance simplifies toEach summand is nonnegative; the sum may be infinite. The pairing converges by Cauchy-Schwarz inequality. This definition does not silently assume a subgradient exists at every $v\in {\ell }^1$.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

The cubic amplitude saturation equation $R^{\prime }=R(Q^2-R^2)$ and the Picard-Lindelöf theorem imply that positive solutions cannot cross the equilibrium $Q$. If $0<R_0<Q$, then $R$ stays below $Q$ and increases. If $R_0>Q$, then it stays above $Q$ and decreases. The explicit cubic amplitude saturation solution tends to $Q$ in both cases, soThe same relation makes the distance a decreasing monotone function in any spatial norm for which the plane wave has finite norm, for example on a periodic domain. At large time, $R-Q\sim -\frac{Q}{2}(Q^2/R_0^2-1)e^{-2Q^{2}t}$.

The printed monotonicity wording must be interpreted as monotone amplitude relaxation, or monotone decay of the distance between the waves. Complex numbers have no ordinary monotone ordering, and the real part at a fixed point can oscillate. For instance, $k=U=1$, $\mu =2$ and $R_0=1/2$ give $\mathrm{Re}{\psi }_R(0,t)=R(t)\cos t$, which is not monotone despite the monotone amplitude. This result concerns the common-phase single-mode family, not stability against arbitrary spatial perturbations.

A pointed cone has no nonzero line through the origin: equivalently $K\cap (-K)=\{0\}$. If $A\in K\cap (-K)$, its quadratic form vanishes on the nonnegative orthant. Testing the coordinate vectors gives $A_{ii}=0$. Testing $e_i+e_j$ then givesSince $A$ is a symmetric matrix, every entry is zero. Thus $\boxed{K\cap (-K)=\{0\}}$.

The identity matrix gives an interior point. For a symmetric perturbation $E$ with operator norm $\Vert E{\Vert }_2<1$, the Cauchy-Schwarz inequality givesThis open norm ball around $I$ lies even in the positive semidefinite cone, hence in $K$. Consequently $\boxed{I\in \mathrm{int}K}$ and the interior is nonempty.

More generally, the interior of the copositive cone consists exactly of strictly copositive matrices. Positivity on the compact nonnegative unit sphere has a positive minimum and persists under small perturbations; a zero there is destroyed by an arbitrarily small negative multiple of $I$.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

For $0<\alpha \le 1$, uniform Lipschitz-$\alpha$ means that one constant $M$ satisfies $|f(x)-f(y)|\le M|x-y{|}^{\alpha }$ for every $x,y\in [0,1]$. The resulting Hölder space has normFor the subsequent estimates at general $\alpha >0$, use the higher-order Hölder class convention: write $\alpha =r+\beta$ with $r\ge 0$ an integer and $0<\beta \le 1$, and require $f\in C^r([0,1])$ with $f^{(r)}$ uniformly Lipschitz-$\beta$. ThusAt integer $\alpha =k$, this uses $C^{k-1,1}$; the alternative classical $C^k$ convention is stronger on this compact closed interval and also suffices for the estimates.

For exponents above one, a first-difference inequality alone forces the function to be constant. Partition $[x,y]$ into $n$ equal pieces and use the triangle inequality: $|f(y)-f(x)|\le M|y-x{|}^{\alpha }{n}^{1-\alpha }\to 0$. Consequently a nontrivial higher-order definition is needed for the full range of exponents in the later wavelet estimates. Equivalently, higher-order Hölder class gives a local Taylor polynomial with remainder bounded by $C\Vert f{\Vert }_{C^{\alpha }}|x-x_0{|}^{\alpha }$.

The strict null space property of order $s$ isHere $v_S$ agrees with $v$ on $S$ and is zero elsewhere. Suppose a nonzero null vector had at most $2s$ nonzero entries. Split its support of a vector into disjoint sets $S,T$ of size at most $s$. Applying the null space property to both sets would give $\Vert v_S{\Vert }_1<\Vert v_T{\Vert }_1$ and $\Vert v_T{\Vert }_1<\Vert v_S{\Vert }_1$, a contradiction. Empty parts cause the same contradiction. Thus no such nonzero null vector exists.

For an $s$-sparse $x$, the feasible vector $x$ has $\Vert x{\Vert }_0\le s$, where the zero-subscript quantity counts nonzero entries and is not a norm. Any feasible competitor $z$ with $\Vert z{\Vert }_0\le \Vert x{\Vert }_0$ also has at most $s$ nonzero entries. The difference $z-x\in \ker A$ has at most $2s$ nonzero entries, so $z=x$. Competitors with larger support have strictly larger objective. Therefore every s-sparse vector is the unique sparsest feasible vector. This is sparse injectivity; for $s=0$ the only sparse vector is zero and the conclusion is immediate.

Absolute values are missing in the printed formula. For signed real blocks, the intended block mixed norm must beWithout them, at $q=1$ a block $(-1,1)$ has value zero although it is nonzero, and a negative scalar has negative value. For nonintegral $q$, a negative component may not even have a real power. The literal expression therefore is not a norm and does not define the claimed general convex problem. Even integer exponents do not fix all other $q$ in the stated range.

With the corrected norm, let $q^{\prime }$ be the Holder conjugate exponent: $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Blockwise Holder inequality, with equality from a norming block, givesThe duality of block mixed norms identifies the closed compact convex set and the projection residual from the Moreau decompositionFor $q=2$, the dual blocks are Euclidean balls; for $q=1$, they are cubes given by componentwise bounds $|p_{i,j}^r|\le \lambda$. No injectivity or surjectivity of $A$ is required: compactness of the block product makes its linear image compact, and the strictly convex fidelity still gives a unique primal minimizer. For general $q^{\prime }$, the Euclidean projection onto a dual block ball is not generally obtained by radial scaling, so the special clipping formula from the isotropic case should not be reused without further analysis.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincaré inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

A Runge-Kutta method advances $y^{\prime }=f(t,y)$ through stagesThe Butcher tableau records $A,b,c$, normally with $c=Ae$. Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.

For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of $z$ for which the step is defined and $|R(z)|\le 1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z{)}^n{y}_0$: the powers are bounded exactly when $|R(z)|\le 1$. For $y^{\prime }=Ly$, the update is $R(hL)$. If $L$ is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.

An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has $R(z)=1+z$ and a disk of stability $|1+z|\le 1$, so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has $R(z)=1/(1-z)$, and $|1-z|\ge 1$ when $\mathrm{Re}z\le 0$, proving A-stability. It is also L-stable, because $R(z)\to 0$ for large left-half-plane $z$.

The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to $-1$. Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.

For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $G_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Then $D_i=d+h{\sum }_j{a}_{ij}{G}_j$. Expansion of the output squared norm, followed by substituting $d=D_i-h{\sum }_j{a}_{ij}{G}_j$ in its linear terms, proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem $y^{\prime }=\lambda y$, or its two-dimensional real form, also proves A-stability whenever stages are well defined.

The Backward Euler method has $b=A=1$, hence $M=1$; the implicit midpoint rule has $b=1$, $A=1/2$, hence $M=0$. Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous $f$, the stage map is a contraction when $hL{\max }_i{\sum }_j|a_{ij}|<1$, so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.

Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows fromThe unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.

Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most $C{h}^{p+1}$, then the error satisfies $\Vert e_{n+1}\Vert \le \Vert e_n\Vert +C{h}^{p+1}$, hence $\Vert e_n\Vert \le \Vert e_0\Vert +CT{h}^p$ for $nh\le T$. A Lipschitz step bound $1+Kh$ gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.

For a fully discrete linear partial differential equation evolution, write $U^{n+1}=C_{h,k}{U}^n$ in a specified discrete norm. Finite-time stability of a numerical method meanswith $C_T$ independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as $e^{\omega T}$ when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.

For constant coefficients on an infinite or periodic uniform grid, von Neumann stability analysis substitutes the Fourier mode $U_j^n=G(\theta {)}^n{e}^{ij\theta }$. The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, $|G(\theta )|\le 1$ proves contractivity, while $|G(\theta )|\le 1+\omega k$ gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.

For the Forward Euler diffusion scheme with $\mu =k/h^2$,Thus $|G|\le 1$ for all frequencies exactly whenSufficiency follows since $G\in [-1,1]$; necessity follows from the highest-frequency mode $\theta =\pi$ on even periodic grids. This gives the usual parabolic mesh restriction $k\le h^2/2$. For the Backward Euler diffusion scheme,so every $\mu \ge 0$ is stable. The Crank-Nicolson method hasagain of modulus at most one for all $\mu \ge 0$, but stiff modes approach $-1$ rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.

For advection $u_t+a{u}_x=0$, assume $a>0$ and set $\nu =ak/h$. The upwind finite difference scheme is $U_j^{n+1}=(1-\nu )U_j^n+\nu U_{j-1}^n$, withIt is contractive when $0\le \nu \le 1$. This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative $a$, the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has $G=1-i\nu \sin \theta$ and $|G|>1$ for nonzero $\nu \sin \theta$. Under fixed nonzero $\nu$ refinement it is unstable, since a fixed nontrivial mode grows geometrically over $T/k$ steps. This is not a claim of instability under every imaginable mesh coupling: $k=O(h^2)$ instead bounds its spurious finite-time growth, since $\log |G|\le a^2{k}^2/(2h^2)$.