Let $(x_n,y_n)$ be a Cauchy sequence in the product. Since each coordinate distance is bounded by $\delta$, both $(x_n)$ and $(y_n)$ are Cauchy in their respective factors. By completeness they have limits $x\in M$, $y\in N$. Part (ii) then gives convergence of the pairs to $(x,y)$.

Thus the product of two complete metric spaces is complete under the maximum product metric. This checks every Cauchy sequence, rather than assuming beforehand that it has a coordinate limit.

Each coordinate distance is at most the maximum, so convergence of the pairs implies convergence of both coordinates. Conversely, for any $ϵ>0$, coordinate convergence gives indices $N_M,N_N$ after which the respective distances are below $ϵ$. Beyond their maximum, both are below $ϵ$, and so is the maximum product metric.The same distance comparison will also handle Cauchy sequences.

Nonnegativity and symmetry follow from the corresponding metric axioms in the two factors. The maximum is zero exactly when both coordinate distances are zero, hence exactly when the ordered pairs are equal.

For the triangle inequality, put $p=(m_1,n_1)$, $q=(m_2,n_2)$ and $r=(m_3,n_3)$. ThenTherefore the maximum formula defines a metric, the maximum product metric.

True. For every finite partial sum,The nondecreasing partial sums on the left are bounded, so the square series converges. Hence ${\ell }^1\subseteq {\ell }^2$ andApplying this norm inequality to $x-y$ proves $\Vert f(x)-f(y){\Vert }_2\le \Vert x-y{\Vert }_1$. Thus the inclusion is a Lipschitz continuous map, in particular continuous, and it is well-defined on every element of l1 sequence space.

True. For any continuous map, $f(\overline{A})\subseteq \overline{f(A)}$: every open neighbourhood of $f(x)$ pulls back to an open neighbourhood of $x\in \overline{A}$, and therefore meets $A$ in its preimage.

Apply the same inclusion to the continuous inverse of the homeomorphism:Applying $f$ gives the reverse inclusion, provingThe closures are taken in their respective ambient spaces.

True. For $u,v\in f(M)$ choose preimages $x,y\in M$. A path $\gamma :[0,1]\to M$ joining $x$ to $y$ exists because $M$ is path-connected. The composition of continuous functions $f\circ \gamma$ is a path in $f(M)$ from $u$ to $v$.Surjectivity onto a larger target is unnecessary because the conclusion concerns the image itself.

False: the target need not equal the image. Take the one-point compact space $M=\{0\}$, $N=\mathbb{R}$, and the continuous map $f(0)=0$. The target is not compact, for example because it is unbounded.

The valid theorem is that a continuous image of a compact space is compact. Thus $f(M)$ is compact; compactness of $N$ follows if surjectivity is added.

False. Removing a point from the real line leaves two disjoint nonempty open rays, so the remainder is disconnected. Removing a point from the Euclidean plane leaves a path-connected space: join any two remaining points by a line segment if it avoids the removed point, or by two segments through a point not on either of the two lines through the removed point and the chosen endpoints.

A proposed homeomorphism would restrict to a homeomorphism between these punctured spaces and preserve connectedness, which is impossible. Hence

The closure $\overline{A}$ is the smallest closed subset of $M$ containing $A$, equivalently the intersection of all such closed subsets. In a metric space it is characterized byIf some ball misses $A$, its centre lies outside a closed superset of $A$; conversely the complement of any closed superset is an open neighbourhood disjoint from $A$. This verifies the equivalence of the two descriptions.

A homeomorphism $f:M\to N$ is a bijection for which both $f$ and $f^{-1}$ are continuous maps, using the respective metric topologies.Continuity of a bijection alone is insufficient: continuity of the inverse is a separate requirement.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

Using the dual pairing between $X^{\ast }$ and $X$, a monotone operator satisfiesA nonlinear coercive operator with respect to $u_0$ satisfiesThe directional derivative is ${\partial }_{v}F(u)={\lim }_{t\to 0}[F(u+tv)-F(u)]/t$, when this finite limit exists through admissible points. At a boundary point of $M$ only an inward direction may be admissible; then one uses the corresponding one-sided limit.

If $A:X\to X^{\ast }$ is monotone and $s<t$, apply its defining inequality at $u+tv$ and $u+sv$. It becomes $(t-s)[\phi (t)-\phi (s)]\ge 0$, so $\phi$ is nondecreasing. Conversely choose $u=y$, $v=x-y$. Nondecrease from zero to one gives $⟨A(x)-A(y),x-y⟩\ge 0$. This proves the line criterion for a monotone operator.

The displayed nonlinearity is singular at $s=-1$, so it cannot be used directly as the globally bounded function in part (b). On $s\ge 0$, however,Use nonnegative truncation of a semilinear source: define $F(s)=f(\max \{s,0\})$. This is continuous and bounded on all of $\mathbb{R}$, with $F\ge 1/2$. Part (b) gives a weak solution of $-\Delta u=F(u)$, $u\in H_0^1$.

The weak maximum principle in the needed form says that if $u\in H_0^1$ and $\int \nabla u\cdot \nabla \phi \ge 0$ for every nonnegative $\phi \in H_0^1$, then $u\ge 0$ almost everywhere. Here this also follows directly by testing with the negative part $u_{-}=\max \{-u,0\}\in H_0^1$:The Poincaré inequality gives $u_{-}=0$. Hence $u\ge 0$, $F(u)=f(u)$, and the denominator $1+u$ stays at least one almost everywhere. Thus the original equation has a nonnegative weak solution, with bounded right side.

A weak solution is $u\in H_0^1(\Omega )$ such thatLet $M_f=\Vert f{\Vert }_{\infty }$ and define the Nemytskii operator $N(v)=f\circ v$ on $L^2(\Omega )$. It satisfies $\Vert N(v){\Vert }_2\le M_f|\Omega {|}^{1/2}$. It is continuous in $L^2$: from any sequence $v_j\to v$ in $L^2$, every subsequence has a further subsequence converging almost everywhere; continuity and boundedness of $f$, followed by the dominated convergence theorem, give convergence of the corresponding $N(v_j)$ in $L^2$. This subsequence criterion proves continuity of the whole sequence.

For $h\in L^2$, the Lax-Milgram theorem defines the inverse Dirichlet solution map $Sh\in H_0^1$ by $\int \nabla (Sh)\cdot \nabla \phi =\int h\phi$. Testing with $Sh$ and using the Poincaré inequality yieldsThe Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. Therefore $T=S\circ N$ is a nonlinear compact operator and its image lies in the closed $L^2$ ball of radius $R=1+C_P^2{M}_f|\Omega {|}^{1/2}$. This ball is nonempty, closed, bounded and convex. The Schauder fixed-point theorem gives $u=Tu$. Since $Tu\in H_0^1$, its defining identity is exactly the required weak formulation. Thus a weak solution exists. This is the bounded semilinear Dirichlet existence theorem.

Here $T$ is a nonlinear compact operator: it is continuous and sends bounded subsets of $M$ to relatively compact subsets of $X$. Equivalently, the images of any bounded input sequence have a norm-convergent subsequence.

The Schauder fixed-point theorem states that if $M$ is a nonempty closed bounded convex set in a Banach space and $T:M\to M$ is continuous and compact, then $\boxed{T\text{ has a fixed point in }M}$. Linearity, a contraction estimate, and uniqueness are not assumptions or conclusions of this theorem.

The space $C([a,b])$ with the supremum norm is a Banach space. Continuity of the kernel makes $K$ a bounded linear operator. Induction on $n$ givesso $A^{n}y-A^{n}z=K^n(y-z)$. The ordered integration region gives the factorial bound for a Volterra iterate:For completeness, if this holds at $n$, then integrating $M{M}^n(s-a{)}^n/n!$ from $a$ to $t$ gives the bound at $n+1$; the initial case is the defining integral. Thusand ${\sum }_{n\ge 1}{\alpha }_n=e^{M(b-a)}-1<\infty$. Part (b) now gives uniform convergence of every iteration sequence to the unique fixed point. Since $y=Ay$ is precisely the Volterra integral equation, this is its unique continuous solution. No restriction $M(b-a)<1$ is needed: the factorial controls all iterates on the whole finite interval.

This is the fixed point theorem for summably Lipschitz iterates. Put $x_n=A^n{x}_0$ and $d=\Vert x_1-x_0\Vert$. For $n\ge 1$,Consequently, for $m>n\ge 1$,Summability makes the tails tend to zero, so $(x_n)$ is a Cauchy sequence. Since $M$ is closed in a Banach space, $x_n\to x\in M$. The $n=1$ hypothesis makes $A$ a Lipschitz map, hence continuous; passing to the limit in $x_{n+1}=A{x}_n$ gives $x=Ax$.

If $y$ is another fixed point, then $\Vert x-y\Vert \le {\alpha }_n\Vert x-y\Vert$ for every $n$. Since ${\alpha }_n\to 0$, eventually ${\alpha }_n<1$ and thus $x=y$. Finally let $m\to \infty$ in the tail estimate:The map itself need not be a contraction mapping; its successive iterates provide the summable error bound.

The Brouwer fixed-point theorem says that every continuous self-map of a closed ball in ${\mathbb{R}}^n$ has a fixed point.

Suppose $g$ has no zero in ${\overline{B}}_R$. Thenis a continuous map from ${\overline{B}}_R$ into its boundary, hence into ${\overline{B}}_R$. The Brouwer fixed-point theorem gives $T(x_0)=x_0$. This forces $|x_0|=R$ andcontradicting the assumed boundary inequality. Therefore $\boxed{g(x_0)=0\text{ for some }|x_0|\le R}$. This is the outward vector field zero theorem.