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The contraction mapping theorem states that a contraction mapping of a nonempty complete metric space into itself has a unique fixed point $x_{\ast }$. For any starting point $x_0$, the iterates $x_{n+1}=f(x_n)$ converge to $x_{\ast }$, with the quantitative estimateCompleteness and the self-map condition are essential: the geometric estimate makes the iterates a Cauchy sequence, completeness supplies their limit in $X$, and continuity of the contraction mapping makes that limit a fixed point. If two fixed points existed, their distance would be at most $q$ times itself, forcing equality of the points.

Define $F_i(x)=a_i+\frac{1}{6}{\sum }_{j\ne i}\sin x_j$. Solving the system is equivalent to finding a fixed point of $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$. The mean value theorem implies $|\sin u-\sin v|\le |u-v|$. In the specified metric,Each coordinate difference occurs exactly twice. This metric makes ${\mathbb{R}}^3$ a nonempty complete metric space, because a Cauchy sequence in it is Cauchy in each real coordinate, and coordinate convergence implies convergence in the sum metric. Thus $F$ is a contraction mapping with constant $1/3$.

There is exactly one solution for every choice of $(a_1,a_2,a_3)$, by the contraction mapping theorem. Iteration of $F$ from any starting vector also constructs it.

Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

A contraction mapping satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one fixed $0\le q<1$. The contraction mapping theorem states that such a map on a nonempty complete metric space has a unique fixed point and that every sequence of iterates converges to it.

Choose $x_0$ and put $x_{n+1}=T{x}_n$. Iteration gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$,Thus the iterates form a Cauchy sequence and converge by completeness to $x_{\ast }$. The Lipschitz bound gives continuity of $T$, hence $T{x}_{\ast }=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing $m\to \infty$ also gives the usual geometric error estimate. For $q=0$, convergence occurs after one application.

If $f^k$ is a contraction mapping, let $x_{\ast }$ be its unique fixed point. The commutation identity $f^{k}f=f{f}^k$ gives $f^k(f{x}_{\ast })=f{x}_{\ast }$, so uniqueness implies $f{x}_{\ast }=x_{\ast }$. Any fixed point of $f$ is a fixed point of $f^k$, giving uniqueness, without even assuming $f$ continuous. On the complete discrete metric space $\{0,1,2\}$, the map $f(0)=f(1)=0$, $f(2)=1$ has constant second iterate but is not a contraction: $d(f2,f0)=d(2,0)=1$.

For the specified function, $x+\sin x\ge 0$ on $[0,\infty )$ because it starts at zero and has derivative $1+\cos x\ge 0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfiesThe mean value theorem therefore makes it a contraction of constant at most $2/3$. Hence

Choose $x\in X$ and iterate $x_{n+1}=f(x_n)$. The contraction mapping estimatemakes $(x_n)$ Cauchy by a geometric series. Its limit $x_0$ satisfies $f(x_0)=x_0$ because $f$ is Lipschitz. Two fixed points would satisfy $d(x_0,y_0)\le \lambda d(x_0,y_0)$, so they coincide. This proves the Banach fixed-point theorem.

Now suppose the final assertion failed. There would be $m_k,n_k\to \infty$ with $f(a_{m_k})=a_{n_k}$. Since $a_n\to a$ and $f$ is continuous, the two sides tend respectively to $f(a)$ and $a$, so $f(a)=a$. Uniqueness would give $a=x_0$, a contradiction. Hence the required $\ell$ and $m$ exist.

The Fredholm alternative for an elliptic Dirichlet problem and part (b) give an inverse $S:Y\to X$ of norm at most $K\ge 1$. ChooseOn the closed radius-$\epsilon$ ball $\mathcal{B}\subset X$, set $T(v)=S(\mathcal{Q}(v)+f)$. The stated inequalities giveandThus $T$ is a contraction mapping of the complete metric space $\mathcal{B}$ into itself. The Banach fixed-point theorem gives $u=T(u)$, solving the required nonlinear elliptic boundary value problem. This small-data existence for a nonlinear elliptic Dirichlet problem also gives uniqueness within this small ball; it does not assert global uniqueness.

For $f,g\in X$, the Lipschitz condition givesChoose $c$ so that $c+1>Mb$. Then $\varphi$ is a contraction mapping. The space $X$ is complete in the sup norm, and therefore also in the equivalent weighted norm.

For initial value $y_0$, defineIt has the same contraction constant. The Banach fixed-point theorem gives a unique fixed point $f$, which satisfies the integral equation. The fundamental theorem of calculus then givesConversely every solution satisfies that integral equation, proving uniqueness on $[1,b]$. This weighted-norm argument is the Bielecki norm method.

A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

For $0<\mu <1$, one iterate sends the whole interval into $[1-\mu ,1]$, whereThis is a contraction mapping with unique fixed pointEvery orbit converges to $x_{\ast }$, alternating around it because the multiplier is $-\mu$.

For $\mu =1$, the first iterate lies in $[0,1]$ and $F(x)=1-x$ there. Hence $F^2$ is the identity on $[0,1]$: $1/2$ is fixed, while every other point in that interval belongs to the period-two orbit $\{x,1-x\}$.

The preceding estimate gives constants $A,B>0$ such that, whenever $\Vert w{\Vert }_{H^2}\le R$,Choose $R>0$ so that $B{R}^2\le 1/2$, and then choose $r>0$ so that $Ar\le R/2$. If $\Vert f{\Vert }_2\le r$, the closed ball $B_R(0)$ is mapped into itself.

For $w,z\in B_R(0)$, factor the cubic gradient nonlinearity asThe same Sobolev embedding theorem and the Holder inequality implyThe elliptic estimate therefore yieldsShrinking $R$ further makes $C^{\prime }{R}^2<1$, so $\Phi$ is a contraction mapping of the closed ball. This ball is complete because $H^2(U)\cap H_0^1(U)$ is a Banach space. The contraction mapping theorem gives a fixed point $u=\Phi (u)$, and its defining equation isThus the nonlinear elliptic boundary value problem has a solution for sufficiently small $\Vert f{\Vert }_2$.

Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Suppose the inverse $S:C^{0,\alpha }\to C_0^{2,\alpha }$ of a linear elliptic boundary value problem is bounded by $K$, and a nonlinear map satisfiesThen $T(v)=S(Q(v)+f)$ maps a sufficiently small closed ball into itself and is a contraction mapping when $\delta$ and $\Vert f{\Vert }_{C^{0,\alpha }}$ are sufficiently small. The Banach fixed-point theorem gives a solution of $Lu=Q(u)+f$ with zero Dirichlet boundary condition, unique within that ball. Invertibility follows from the Fredholm alternative for an elliptic Dirichlet problem when the homogeneous null space is zero.