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Write $f(u)=a{u}^2/(b^2+u^2)$, with $b>0$. Besides zero, the fixed points satisfy $u^2-au+b^2=0$. If $a^2>4b^2$, they areThe derivative is $f^{\prime }(u)=2a{b}^{2}u/(b^2+u^2{)}^2$. At a positive fixed point, $b^2+u^2=au$, so $f^{\prime }(u)=2b^2/(au)=2(1-u/a)$. Thus zero is attracting, $u_{-}$ is repelling because $f^{\prime }(u_{-})>1$, and $u_{+}$ is attracting because $0<f^{\prime }(u_{+})<1$.

Moreover,For $0<u_0<u_{-}$ the iterates strictly decrease and stay positive. Their limit is a fixed point below $u_{-}$, hence zero. For $u_{-}<u_0<u_{+}$ the iterates increase towards $u_{+}$; for $u_0>u_{+}$ they decrease towards $u_{+}$. Monotonicity of $f$ prevents crossing the limiting fixed point. $u_{-}$ is the extinction threshold. The cobweb plot illustrates these alternatives.

For $a^2<4b^2$ there are no positive fixed points, and $f(u)<u$ for every $u>0$, so all populations tend to zero. At $a^2=4b^2$, the positive fixed point $a/2$ is semistable: initial populations below it tend to zero, and those above it decrease towards it. This is a saddle-node bifurcation in the discrete dynamics.

A control policy specifies each control as a nonanticipating measurable function of the observed history, possibly with additional randomization. It may depend on time and the entire past; a stationary Markov policy uses only the current state. We use the standard admissibility assumptions allowing measurable choices and concatenation of continuation policies.

The dynamic programming operator is monotone: $f\le g$ implies $\mathcal{L}f\le \mathcal{L}g$. Nonnegative costs give $F_1\ge F_0=0$, so induction gives $F_{t+1}\ge F_t$. Thus $F_t$ increases pointwise to an extended nonnegative limit $F_{\infty }$. Backward induction identifies $F_t$ with the infimum of expected costs for the first $t$ periods: condition on the first control and next state, then optimize the continuation. Every infinite-horizon policy has at least that finite-horizon cost, so

For the Bellman equation, any infinite-horizon policy, conditional on its first action and next state, has continuation cost at least $F(X_1)$. Its total cost is therefore at least $\mathcal{L}F(x)$; infimizing gives $F\ge \mathcal{L}F$. For the reverse inequality at a state with finite $\mathcal{L}F(x)$, choose a first action within $\epsilon /2$ of its infimum and concatenate continuation policies whose costs are within $\epsilon /(2\beta )$ of $F(y)$ at each next state $y$. This produces a policy of cost at most $\mathcal{L}F(x)+\epsilon$. Such continuations are needed only at states of finite value, which have probability one under an action with finite continuation expectation. Let $\epsilon \downarrow 0$. If $\mathcal{L}F(x)=\infty$, the first inequality already gives $F(x)=\infty$. Therefore

Let $\Phi =\mathcal{L}\Phi \ge 0$, and use the permitted selector $u_{\ast }$. Under the stationary policy $u_{\ast }(X_t)$, repeated conditional expectation givesIf $\Phi (x)$ is infinite the desired comparison is immediate; otherwise these nonnegative quantities have finite expectations and the displayed recursion is legitimate. Drop the terminal term and apply monotone convergence theorem as $T\to \infty$. The policy's infinite-horizon cost is at most $\Phi (x)$, soThus the value is the least nonnegative Bellman fixed point. The argument works also for $\beta =1$ because costs and the dropped terminal term are nonnegative; it makes no unwarranted contraction or finite-value assumption.

Use binary expansions ending in zeros at dyadic rationals. The doubling map shifts the expansion left:In ordinary coordinates it is $2x$ on $[0,1/2)$ and $2x-1$ on $[1/2,1)$. The branches are continuous; the left limit at $1/2$ is $1$ while its value and right limit are $0$. This is its only discontinuity in the half-open interval.

For every $N\ge 2$, the repeating binary block $0^{N-1}1$ givesIts least period is $N$: a shorter repeating block would make more than one digit equal to one in a length-$N$ period. The orbit avoids the discontinuity, and its multiplier is $(F^N{)}^{\prime }=2^N>1$, so these periodic orbits are repelling and unstable. The same multiplier argument applies to every periodic orbit, with the endpoint fixed point treated one-sidedly.

In the two-branch horseshoe formulation of Glendinning chaos, disjoint subintervals are each mapped onto the same interval by an iterate. Here $J=(0,1)$, $K_0=(0,1/2)$ and $K_1=(1/2,1)$ already satisfy $F(K_0)=F(K_1)=J$. Their inverse branches $x\mapsto x/2$ and $x\mapsto (x+1)/2$ realize every binary itinerary. This is a horseshoe for a piecewise continuous map; on the circle the endpoint jump disappears. The symbolic explanation also shows dense periodic points, by repeating any prescribed finite prefix, and topological transitivity, because an iterate maps any sufficiently small dyadic cylinder onto $(0,1)$. Finally every neighborhood contains points sharing an arbitrarily long prefix but with different subsequent digits; an appropriate iterate separates their images by a fixed amount, for instance more than $1/4$. This proves sensitive dependence on initial conditions as well. The binary shift has full two-symbol chaotic dynamics.

No. Define $f(x)=1$ for $0\le x<1$ and $f(1)=0$. This self-map of $[0,1]$ is discontinuous at one. Every point smaller than one is sent to a different point, and one is sent to zero, so it has no fixed point.

No. The continuous function $f(x)=x+1$ is a self-map of $\mathbb{R}$, but the fixed point equation would require $x+1=x$, which is impossible.

No. The continuous function $f(x)=(x+1)/2$ maps $(0,1)$ into $(0,1)$, but $f(x)-x=(1-x)/2>0$ throughout the domain. Its only possible fixed point is the excluded endpoint one.

Yes: a tangential zero and a crossing zero can give exactly two fixed points. One explicit continuous piecewise linear example isAdjacent formulas agree at each junction. Their endpoint values lie in $[0,1]$, so linear interpolation proves that the whole image lies in $[0,1]$; also $f(0)=1$ and $f(1)=0$. On the four pieces, $f(x)-x$ is respectively $1-3x$, $3x/2-1/2$, $1-3x/2$, and $2-3x$. Their only zeros on their respective intervals are $1/3$ and $2/3$. Thus the complete fixed point set is $\boxed{\{1/3,2/3\}}$. Continuity forces at least one zero, but does not force every zero to change sign.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous, then every real value between $f(a)$ and $f(b)$ is attained at some $c\in [a,b]$. Here is a proof using completeness of the real numbers. Endpoint values are immediate, so suppose $f(a)<y<f(b)$; reversing the sign of $f$ handles the other ordering. LetThe set is nonempty and bounded above. Continuity at $a$ puts points to the right of $a$ in $S$, while continuity at $b$ excludes an entire interval to the left of $b$, so $a<c<b$. If $f(c)<y$, continuity puts some point larger than $c$ in $S$, contradicting its supremum. If $f(c)>y$, continuity excludes $S$ in a neighborhood of $c$, contradicting the existence of points of $S$ arbitrarily close to its supremum from below. Therefore $f(c)=y$, proving the theorem.

For the fixed-point assertion, put $g(x)=f(x)-x$. This is continuous, with $g(0)=f(0)\ge 0$ and $g(1)=f(1)-1\le 0$. If either endpoint value is zero it already gives a fixed point; otherwise the intermediate value theorem gives an interior zero. HenceThe proof needs both continuity and the closed-interval endpoint information; the following counterexamples separate those hypotheses.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.

The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. The Jacobian matrix isAt the origin its eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle equilibrium for every allowed $\mu$. At either nonzero fixed point the characteristic polynomial is ${\lambda }^2+\lambda +2{\mu }^2$, givingThus both are stable nodes for $0<|\mu |<1/(2\sqrt{2})$ and stable focuses for $|\mu |>1/(2\sqrt{2})$. At equality the repeated eigenvalue is $-1/2$ and the matrix is not scalar, so there is only one independent eigenvector: the fixed point is a stable improper node.

The stable manifold of the origin is $x=0$, exactly: there $\dot{y}=-y$. Conversely, a trajectory converging to the origin with $x\ne 0$ would eventually have $\dot{x}/x={\mu }^2-y>{\mu }^2/2$, which precludes convergence of $x$ to zero.

For the local unstable manifold, write $y=h(x)=a{x}^2+b{x}^4+O(x^6)$. Reflection symmetry makes $h$ even, and the invariance equation for a graph requiresComparing quadratic and quartic terms gives $2a{\mu }^2=1-a$ and $4b{\mu }^2-2a^2=-b$. Therefore

For $\lambda >1$, the multiplier at the origin exceeds one, so it is unstable. The other fixed points areUsing fixed point stability for an iteration, $|3-2\lambda |<1$ exactly when $1<\lambda <2$, and $|3-2\lambda |>1$ when $\lambda >2$. HenceThe endpoint $\lambda =2$ has multiplier $-1$ and requires nonlinear analysis, which is not part of the asserted open intervals.

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.

The preceding summer supplies $\gamma M_n$ new seeds. After one winter, $\alpha \gamma M_n$ survive and the fraction $\mu$ germinates. Seeds from summer $n-1$ must survive two winters and fail to germinate after the first, so their germinating contribution is $\nu {\alpha }^2(1-\mu )\gamma M_{n-1}$. Older seeds cannot contribute. Applying the density-dependent maturation function givesThis is a seed-bank population recurrence. Put $A=a+b$. Its fixed points satisfy $M_{\ast }=N(A{M}_{\ast })$. The zero solution always exists. If $A>1$, the graph of $N(AM)$ leaves the origin with slope $A>1$, is concave because $N^{\prime }$ decreases, and is bounded above by $N_{\max }$; it therefore crosses the straight line $M$ again at a positive value. Concavity makes this positive crossing unique: the ratio $N(S)/S$ is nonincreasing, starts at one, tends to zero, and is strictly decreasing at any level below one that it attains after the initial slope has dropped. Thus there are exactly two nonnegative steady states when $A>1$.

Linearization at extinction uses $N^{\prime }(0)=1$:The positive root is $r_{+}=(a+\sqrt{a^2+4b})/2$. Since the polynomial at $r=1$ is $1-A<0$, and tends to positive infinity, $r_{+}>1$. Its associated two-component perturbation can be chosen nonnegative, so the growing mode is physically admissible. Extinction is unstable whenever $\boxed{a+b>1}$.

Use the discontinuous exampleIn both pieces $0<f(x)<x$. For $x>1/2$, the image stays strictly above $1/2$, and induction givesThis positive limiting point is not a fixed point, since $f(1/2)=1/4$. The discontinuity is precisely why the limit cannot be passed through $f$. Therefore the pointwise zero-limit conclusion can fail without continuity.