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The Browder convergence theorem for averaged operators states: if $T:{\mathbb{R}}^d\to {\mathbb{R}}^d$ is an averaged operator and its fixed-point set $\mathrm{Fix}T=\{z:Tz=z\}$ is nonempty, then for every starting point, $z^{k+1}=T{z}^k$ converges in norm to a fixed point of $T$. The theorem assumes a fixed point exists; averaging by itself does not create one.

The finite-dimensional mechanism is worth making explicit. For any fixed point $p$, the averaged-operator inequality givesThis is Fejér monotonicity with respect to the fixed-point set. It makes the sequence bounded and, by summation, makes ${\sum }_k\Vert z^{k+1}-z^k{\Vert }_2^2$ finite. Hence its residual tends to zero. Choose a convergent subsequence $z^{k_j}\to \bar{z}$; continuity of the nonexpansive mapping $T$ implies $T\bar{z}=\bar{z}$. Apply Fejér monotonicity with $p=\bar{z}$. The nonincreasing distances $\Vert z^k-\bar{z}{\Vert }_2$ have a subsequence tending to zero, so the whole sequence converges to $\bar{z}$. This last compactness step uses finite dimension; a corresponding general Hilbert space statement usually gives weak convergence.

Assume $\mathrm{argmin}f\ne \varnothing$, as is necessary for convergence to a minimizer. The printed hypotheses alone omit this condition: $f(x,y)=x$ in scalar $x$ is convex with an $L$-Lipschitz gradient for any prescribed $L>0$, but has no minimizer, and the update sends $x^k=x^0-k\tau$ to $-\infty$.

Under existence, define the full update $T(x,y)=(x^{+},y^{+})$. The proximal $y$-step is uniquely defined by strong convexity of its quadratic term and satisfiesThe partial subdifferential in this equation belongs to the scalar function $y\mapsto f(x,y)$; differentiating a subdifferential itself would not define the printed proximal minimization.

We prove directly that the alternating proximal-gradient operator $T$ is averaged. For two inputs $z=(x,y)$ and $\tilde{z}=(\tilde{x},\tilde{y})$, setThe input difference is $(p,q+\tau b)$, the output difference is $(p-\tau a,q)$, and the difference of the two update residuals is $\tau (a,b)$. Exact expansion givesThe first inequality uses cocoercivity at the mixed points $w,\tilde{w}$. Therefore $T$ is $(\tau L/2)$-averaged for the requested $0<\tau L\le 1$. In particular, the endpoint $\tau L=1$ is covered: $T$ is firmly nonexpansive, not merely nonexpansive.

A fixed point has $y^{+}=y$ and $x^{+}=x$, so the two optimality equations give ${\nabla }_{y}f=0$ and ${\nabla }_{x}f=0$. Conversely, at a minimizer both gradients vanish and neither step moves. By convexity,The assumed existence of a minimizer supplies a fixed point. The Browder convergence theorem for averaged operators now givesNo uniqueness of the minimizer is needed. The same estimate even proves averagedness for $0<\tau L<2$ in this smooth setting, but the requested interval already suffices. The existence condition and the use of the gradient at $(x^k,y^{k+1})$ are both essential to the proof as presented.

Use the product topology on $\mathcal{C}=[k{]}^{\mathbb{Z}}$, with $[k]$ discrete, and the left shift $(\mathcal{L}c)(j)=c(j+1)$. A basic cylinder set specifies finitely many coordinates. This is a compact metric space, and $\mathcal{L}$ is a homeomorphism. Write $Y=\overline{\{{\mathcal{L}}^{n}c:n\ge 0\}}$ for the forward orbit closure. A minimal point means that $Y$ is a minimal dynamical system, equivalently that every point of $Y$ has a dense forward orbit in $Y$; it need not be a fixed point.

There is a convention needed in the source: the bounded-gaps property concerns every finite integer interval $I$, hence every finite word over an alphabet. Literally allowing $I=\mathbb{Z}$ would make the displayed condition impossible for finite $U$, although a constant coloring is a minimal point. Under the standard finite-word interpretation, the property is precisely uniform recurrence.

Suppose first that $Y$ is a minimal dynamical system. Its nonempty compact forward-invariant subset $\mathcal{L}(Y)$ must equal $Y$. As the ambient left shift is invertible, all integer translates of $c$ belong to $Y$. Let $I=[a,b]\cap \mathbb{Z}$ have length $\ell =b-a+1$, and let $V=\{z\in Y:z{|}_I=c{|}_I\}$. This is a nonempty clopen set. Each forward orbit in $Y$ meets $V$, so $\{{\mathcal{L}}^{-n}V:n\ge 0\}$ covers $Y$. By compactness, finitely many suffice; let $D$ be the largest index in this finite cover. For any integer $u$, the point ${\mathcal{L}}^{u-a}c$ lies in $Y$, so some $0\le n\le D$ satisfies ${\mathcal{L}}^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell -1]$, entirely inside $[u,u+D+\ell -1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.

Conversely suppose $c$ is uniformly recurrent. Every finite word from any integer translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete alphabet forces eventual exact agreement on that block. Given $I=[a,b]\cap \mathbb{Z}$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\ge 0$, so ${\mathcal{L}}^{n}z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap \mathbb{Z}$ for arbitrarily large $r$ proves that $c$ belongs to the forward orbit closure of every $z\in Y$. That closure is a closed forward-invariant subset of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward orbit is therefore dense in $Y$, provingThe same conclusion holds if orbit closure is defined using all integer iterates: the bounded-gaps condition makes the forward and two-sided closures equal.

The equilibrium equations are $x(-1+x^2+\beta y^2)=y(-1+\beta x^2+y^2)=0$. For $\beta \ne 1$ the fixed points are the origin, four axis points $(\pm 1,0),(0,\pm 1)$, and four diagonal points $(\pm s,\pm s)$ with independent signs and $s=(1+\beta {)}^{-1/2}$. The hypothesis $\beta >-1$ ensures that $s$ is real. The Jacobian matrix has diagonal entries $-1+3x^2+\beta y^2$, $-1+\beta x^2+3y^2$ and off-diagonal entries $2\beta xy$.

At the origin both eigenvalues are $-1$, so it is a stable node. At each axis point the eigenvalues are $2,\beta -1$. At each diagonal point they are $2,2(1-\beta )/(1+\beta )$, along the radial and transverse diagonal directions respectively. ThusThe axes and diagonals are invariant lines. For $\beta =1/2$ the axis saddle points have stable directions transverse to their axes and unstable directions along them; the diagonal points are unstable nodes. For $\beta =2$ these roles switch: the diagonal saddle points have stable transverse directions and unstable radial directions, and the axis points are unstable nodes. Their stable manifolds separate trajectories attracted to the origin from outward-escaping trajectories. Symmetry gives the other quadrants.

For $\beta =1$, the entire unit circle consists of equilibria. In polar coordinates $\dot{r}=r(r^2-1)$ and $\dot{\theta }=0$: every ray inside the circle moves toward the origin, and every ray outside moves outward. The circle has neutral tangential and unstable radial directions; its points are not isolated saddle points or unstable nodes.

All three phase portraits are shown below. The potential $\Phi =-(x^2+y^2)/2+(x^4+y^4)/4+\beta x^2{y}^2/2$ satisfies $(\dot{x},\dot{y})=\nabla \Phi$ and $\dot{\Phi }=|\nabla \Phi {|}^2$, ruling out nonconstant periodic orbits. Blue curves show the saddle stable separatrices, integrated backwards from their local stable directions. The flow arrows and invariant lines, together with the equilibrium classification, specify the topology of the sketches.

A Lie point symmetry is a local one-parameter family of diffeomorphisms of $(t,\mathbf{x})$ that takes solution graphs of the differential equation to solution graphs wherever the transformed graph remains a graph over time. Its infinitesimal generator is $Y=\tau (t,\mathbf{x}){\partial }_t+{\sum }_i{\eta }_i(t,\mathbf{x}){\partial }_{x_i}$. Its first prolongation isFor a regular first-order equation the infinitesimal criterion is ${\mathrm{pr}}^{(1)}Y(\Delta )=0$ on $\Delta =0$; it says that the prolonged vector field is tangent to the equation submanifold.

With the constant nondegenerate canonical matrix $J$, define the Poisson bracket by $\{f,g\}=(\nabla f{)}^{T}J\nabla g$ and the Hamiltonian vector field by $V_f=J\nabla f$. Our convention gives $V_f[h]=\{h,f\}$. Applying the Jacobi identity to a test function $h$,soA first integral $F$ obeys $dF/dt=\{F,H\}=0$. Consequently $[V_F,V_H]=0$, and the local flows commute. The time-independent transformation generated by $V_F$, with $t$ unchanged, therefore maps integral curves of a vector field for $V_H$ to integral curves of a vector field: it is a Lie point symmetry.

Conversely, such a time-independent symmetry gives $[V_F,V_H]=0$, hence $V_{\{F,H\}}=0$. Nondegeneracy of $J$ makes $\{F,H\}$ constant on each connected component. At a fixed point ${\mathbf{x}}_0$, $J\nabla H({\mathbf{x}}_0)=0$ implies $\nabla H({\mathbf{x}}_0)=0$, so that constant is zero on its component. Thus the converse holds on a connected phase space containing a fixed point, or on each component that contains one.

The printed global converse needs this connectedness convention: a fixed point on one component does not control the constants on other components. For example take one component with $H=(q^2+p^2)/2$, $F=0$, and another with $H=p$, $F=q$. The Hamiltonian vector fields commute everywhere and there is a fixed point on the first component, but $F$ is not a first integral on the second. Even on a connected space the fixed-point condition matters: $H=p$, $F=q$ commute but $\dot{F}=1$.

The Householder-John theorem states that if $A=M-N$ is Hermitian positive definite and $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible and $\rho (M^{-1}N)<1$. The associated stationary iteration $x^{(k+1)}=M^{-1}N{x}^{(k)}+M^{-1}b$ therefore converges from every starting vector. For designing a method, choose a splitting with easily solved $M$ and verify the positive-definiteness condition on $M^{\ast }+N$; it is a sufficient certificate and does not require computing every iteration-matrix eigenvalue.

For completeness, put $C=M^{\ast }+N=M^{\ast }+M-A$. If $Mv=0$ for $v\ne 0$, then $v^{\ast }Cv=-v^{\ast }Av<0$, a contradiction. If $M^{-1}Nv=zv$, then $Av=(1-z)Mv$. The case $z=1$ contradicts invertibility of $A$. Taking scalar products and using Hermitian symmetry giveshence $|z|<1$. The usual spectral-radius convergence criterion now proves the stated conclusion.

For the given scheme take the matrix splittingSince $A$ is real symmetric, $U=L^T$, and since it is positive definite, every diagonal entry in $D$ is positive. Thusis positive definite exactly for $0<\omega <2$. The theorem applies, and the iteration matrix has spectral radius below one. The unique solution $x_{\ast }=A^{-1}b$ is its fixed point, so $x^{(k)}-x_{\ast }=(M^{-1}N{)}^k(x^{(0)}-x_{\ast })\to 0$. ThereforeThe scheme is successive over-relaxation; at $\omega =1$ it becomes the Gauss-Seidel method.

The stochastic objective is interpreted as its expectation; a literal pathwise minimization is generally impossible without knowing future noises. Let the value function from time $t$ be $V_t(x)=P_t{x}^2+d_t$, with $P_T=1,d_T=0$. Independence from past observations and zero contemporaneous covariance giveThe Bellman equation is $V_t(x)=x^2+{\min }_u\{c{u}^2+P_{t+1}{V}_{\xi }(Ax+u{)}^2+P_{t+1}{V}_{ϵ}+d_{t+1}\}$. Completing the square yieldsThis feedback is optimal among all controls measurable from the specified past, not merely linear controls. It depends on $V_{\xi }$ and does not depend on $V_{ϵ}$, though the minimal cost does. There is no cost-relevant $u_T$ in the displayed objective.

The backward Riccati recurrence iterates $F(P)=1+A^{2}c{V}_{\xi }P/(c+V_{\xi }P)$ from $1$. It is increasing, maps into $[1,1+A^{2}c]$, and has exactly one positive fixed point. Consequently it converges toAs $T\to \infty$, the finite-horizon feedback at any fixed time tends to this feedback. The closed-loop multiplier is $Ac/(c+V_{\xi }P)$, and its squared-mean factor $\alpha =V_{\xi }{A}^2{c}^2/(c+V_{\xi }P{)}^2$ is less than one, since $(P-1)/P=A^{2}c{V}_{\xi }/(c+V_{\xi }P)$ and $\alpha =(P-1)c/[P(c+V_{\xi }P)]$. Thus the second moment tends to $V_{ϵ}/(1-\alpha )$. The finite-horizon value has $d_1=V_{ϵ}{\sum }_{t=1}^{T-1}{P}_{t+1}$; its average converges by Cesaro summation. For a fixed initial state of finite second moment,The same average is attained by the stationary feedback, by the stationary Bellman equation and bounded second moments.

Let $D$ be the closed unit disc and setThe displacement bound gives $\Vert f(x)\Vert \le 1$, so $f:D\to D$ is continuous. The Brouwer fixed-point theorem applies because $D$ is nonempty, compact and convex. At a fixed point $x_{\ast }$, $x_{\ast }=x_{\ast }-g(K{x}_{\ast })/K$. Therefore $\boxed{t=K{x}_{\ast },g(t)=0}$.

The fixed point zero has multiplier $F^{\prime }(0)=\mu$, so it is attracting for $|\mu |<1$ and repelling for $|\mu |>1$. Other real fixed points satisfyTheir multiplier is $M=1+bx+2a{x}^2$, and they are attracting exactly when $-2<bx+2a{x}^2<0$. At equality $M=\pm 1$ the linear test is inconclusive. For $b\ne 0$ two branches meet at $x_s=-b/(2a)$, ${\mu }_s=1+b^2/(4a)$, a generic saddle-node bifurcation. At $\mu =1$, the nonzero branch crossing zero gives a transcritical bifurcation when $b\ne 0$.

The three sketches use $(a,b)=(1,0),(1,1/2),(-1,-1/2)$ respectively; case (iii) therefore illustrates the subcritical-flip choice.

For $F(x)=x(\mu -x^2)$, the fixed points are zero for all $\mu$ and $\pm \sqrt{\mu -1}$ for $\mu >1$. Their multipliers are respectively $\mu$ and $3-2\mu$, giving attraction for $-1<\mu <1$ and $1<\mu <2$ respectively.

Remove these fixed points from the roots of $F^2(x)=x$. The factor $\mu +1-x^2$ gives the genuine two-cycleIts multiplier is $(2\mu +3{)}^2>1$, so it is always unstable. It collapses to zero at $\mu =-1$, the subcritical period-doubling bifurcation of zero.

The remaining equation is $x^4-\mu x^2+1=0$, whose positive roots in $x^2$ exist only for $\mu \ge 2$. At $\mu =2$ they give the fixed points $\pm 1$, not genuine two-cycles. For $\mu >2$, put $r=\sqrt{(\mu +\sqrt{{\mu }^2-4})/2}>1$. Since $\mu =r^2+r^{-2}$, $F(r)=r^{-1}$ and $F(r^{-1})=r$. There are exactly two further cycles,Their common multiplier is $(\mu -3r^2)(\mu -3r^{-2})=9-2{\mu }^2$. They are attracting for $2<\mu <\sqrt{5}$ and repelling for $\mu >\sqrt{5}$, with marginal multiplier $-1$ at $\sqrt{5}$. Thus there are no cycles for $\mu \le -1$, one for $-1<\mu \le 2$, and three for $\mu >2$.

The nonzero fixed points emerge in a supercritical pitchfork bifurcation at one and each flips to an attracting two-cycle at two. Beyond $\sqrt{5}$ further period doubling to four-cycles is expected, followed by additional bifurcations and potentially chaotic dynamics. This local expectation does not assert chaos for every larger parameter; periodic windows and escape are possible.

Let $f(u)=au(1-u^2)$. A fixed point satisfies $u=f(u)$, soThe nonzero fixed points exist for $a<0$ or $a>1$. Local stability requires $|f^{\prime }(u_{\ast })|<1$, andAway from the boundary values:

At $a=-1,1,2$ a multiplier has modulus one and linear stability is inconclusive; at $a=0$, $0$ is strongly stable.

Consider a triangle of three links, each with capacity $C$. Calls for each pair of its three nodes arrive as independent Poisson processes at rate $\lambda$ and have independent holding times with an exponential distribution of mean one. Try the direct link first; if it is full, try the two-link path through the third node; if either alternative link is full, reject the call. This is alternative routing in a loss network.

In the symmetric Erlang fixed point approximation, let $B$ be each link's blocking probability. A given link receives direct offered traffic $\lambda$. Each of the other two call types contributes overflow traffic $\lambda B$, screened by the availability $1-B$ of the other link on its alternative path. ThusThe extra term expresses a feedback: blocking sends calls onto longer paths, which consume more total capacity and can create still more blocking.

Here is an explicit finite-capacity example, not just a limiting argument. Take $C=1000$ and $\lambda =950$, and let $F(B)=E(950[1+2B(1-B)],1000)-B$. Evaluating the Erlang loss formula givesThe displayed decimals are rounded, but the four signs can be checked exactly using the rational recursion $E(a,0)=1$, $E(a,k)=aE(a,k-1)/(k+aE(a,k-1))$. By the intermediate value theorem, there is a fixed point in each of $(0,0.01)$, $(0.01,0.1)$, and $(0.1,0.4)$. Numerically these three areThus alternative routing can produce multiple Erlang fixed points. This does not imply multiple stationary distributions for the exact finite continuous-time Markov chain, which is irreducible and has a unique stationary distribution; the multiplicity belongs to the approximation. The symmetric alternative-routing model is also discussed in Kelly's review of fixed point models of loss networks.