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A rational lower bound for the natural logarithm proves positivity for every $n$, including $n=1$. DefineThen $h(0)=0$, while direct differentiation givesThus $h(t)>0$ for $t>0$. Setting $t=1/n$ in the expression from part (b) givesTherefore $\boxed{u_n>0\text{ for every }n\ge 1}$, and $(a_n)$ is a strictly increasing sequence.

Along the first sequence, $\sin (1/x_n)=\sin (2n\pi +\pi /6)=1/2$, whereas along the second sequence, $\sin (1/x_n^{\prime })=\sin (2n\pi )=0$. The outer sine therefore givesBoth sequences approach zero through positive values. For arbitrary $a$, the same calculation gives $f_a(x_n)=x_n^a$ and $f_a(x_n^{\prime })=0$.

The difference quotient at zero isFor $a>1$, its absolute value is at most $|h{|}^{a-1}$. Given $\epsilon >0$, set $\delta ={\epsilon }^{1/(a-1)}$. Then $0<|h|<\delta$ makes the difference quotient have absolute value below $\epsilon$. Thus $f_a^{\prime }(0)=0$.

Conversely, along the second sequence the difference quotient is exactly zero. Any finite derivative would consequently have to be zero. Along the first positive sequence, however, the difference quotient equals $x_n^{a-1}$. It is one for $a=1$ and tends to $+\infty$ for $a<1$. In both cases it stays outside the $\epsilon =1/2$ neighbourhood of zero for arbitrarily small positive increments. Thus no finite derivative exists. We obtainIn particular, $0<a\le 1$ gives continuity without differentiability.

For $a>0$, the fact that sine is a bounded function gives $|f_a(x)-f_a(0)|\le |x{|}^a$. Given $\epsilon >0$, choose $\delta ={\epsilon }^{1/a}$. If $|x|<\delta$, then $|f_a(x)-f_a(0)|<\epsilon$. This proves continuity directly from its definition.

For $a\le 0$, use the first sequence from part (b), for which $f_a(x_n)=x_n^a$. When $a=0$ these values are one; when $a<0$ they tend to $+\infty$. In either case, for $\epsilon =1/2$ every neighbourhood of zero contains a sufficiently late $x_n$ with $|f_a(x_n)-f_a(0)|\ge 1/2$. This contradicts the definition of continuity. Hence

Weak convergence $u_j\rightharpoonup u$ means $F(u_j)\to F(u)$ for every $F\in X^{\ast }$. Weak-star convergence $T_j\stackrel{\ast }{\rightharpoonup }T$ means $T_j(x)\to T(x)$ for every $x\in X$, viewing $X$ as the predual. The Sequential Banach-Alaoglu theorem for a separable predual says that every norm-bounded sequence in $X^{\ast }$ has a weak-star convergent subsequence. Separability is what turns the bounded weak-star compactness statement into this sequential form.

For the weak-star limit of periodic rescalings, apply it with $X=L^1(\mathbb{R})$, which is separable, and $X^{\ast }=L^{\infty }(\mathbb{R})$. The rescaled functions have uniform essential-supremum bound $\Vert \theta {\Vert }_{\infty }$. Let $q=\theta -\bar{\theta }$ and $F(t)={\int }_0^{t}q(s)ds$. Since $q$ is one-periodic with zero mean, $F$ is a bounded one-periodic absolutely continuous primitive. Thus for every finite interval,Every weak-star subsequential limit $h$ therefore satisfies ${\int }_a^b(h-\bar{\theta })=0$ for all $a<b$. To apply the supplied uniqueness fact without assuming global integrability, fix $R>0$ and set $v=(h-\bar{\theta })1_{[-R,R]}$. This is in $L^1(\mathbb{R})$, and its integral over any interval is zero, since intersecting with $[-R,R]$ again gives an interval. Hence $v=0$ almost everywhere. Taking integer $R\to \infty$ yields $h=\bar{\theta }$ almost everywhere on the whole line.

Finally, if the full sequence did not converge weak-star to this constant, some $g\in L^1$ and some subsequence would have its pairing with $\theta (jx)-\bar{\theta }$ bounded away from zero. Sequential compactness gives a further weak-star convergent subsequence, whose limit must be the constant just identified, a contradiction. ThereforeThis is averaging against integrable test functions, not a claim of pointwise or strong norm convergence.

Banach's fixed-point theorem states that a map $T:Y\to Y$ on a nonempty complete metric space with $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$ has a unique fixed point, approached from every starting point by iteration.

Choose $x_0$ and let $x_{n+1}=T{x}_n$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. For $m>n$, the triangle inequality and geometric series giveThus the sequence is Cauchy and converges by completeness to some $x_{\ast }$. The contraction estimate implies continuity, so $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$ would force equality of the points. Passing $m\to \infty$ also yields the quantitative error estimate

Brouwer's fixed-point theorem says that every continuous self-map of a nonempty closed Euclidean ball has a fixed point. It does not promise convergence of repeated application. For example $f(x)=1-x$ with starting point $x_1=0$ gives the nonconvergent alternating sequence $0,1,0,1,\dots$, although $1/2$ is a fixed point.

For an invariant ball from sublinear growth, chooseThen $A{R}^{\alpha }\le R/2$ and $B\le R/2$. For every $|x|\le R$ the given estimate yields $|f(x)|\le A{R}^{\alpha }+B\le R$. Thus $f$ restricts to a continuous self-map of the closed ball of this radius, and Brouwer gives a fixed point. The exponent $\alpha <1$ ensures that the linear radius eventually dominates its sublinear growth; no differentiability or contraction estimate is needed.

Using the dual pairing, monotonicity meansStrict monotonicity requires strict positivity whenever $u\ne v$. Coercivity with respect to zero, in the growth sense used for monotone-operator existence theory, meansThis is a condition on all sequences escaping to infinity, not just each fixed direction with a bound depending on that direction. Strict monotonicity concerns pairs of distinct arguments; it does not by itself imply this uniform large-norm growth.

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.

Take an irrational number $a\in (0,1)$. The density of the rational numbers provides a sequence in $\mathbb{Q}\cap [0,1]$ converging to $a$, which is outside that subset. Thus the subset is not closed in $\mathbb{R}$. By the Heine-Borel theorem, it is not compact, despite being bounded.

The sequence $n\in [0,\infty )$ is unbounded. Every compact metric space is bounded, by part (b), so this subspace is not compact.

The points $(n,0)$ belong to the indicated subset for every positive integer $n$. They form an unbounded sequence, so this metric space is not bounded. Therefore it is not compact. Shrinking the vertical width does not bound its horizontal extent.

The sequence converges to $x\in X$ ifEquivalently $d(x_n,x)\to 0$. The limit must belong to the specified metric space, not just to a larger space containing it.

A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence ifThe index $N$ may depend on $\epsilon$, but must work for every pair of sufficiently late terms. No candidate limit is part of this definition.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.

A metric space $(X,d)$ is a complete metric space if every Cauchy sequence in $X$ converges to a point of $X$. Explicitly, the hypothesis isand the conclusion is that some $x\in X$ satisfies $d(x_n,x)\to 0$.

It is a sequentially compact space if every sequence in $X$ has a convergent subsequence whose limit of a sequence belongs to $X$.

It is a connected space if there are no disjoint nonempty open subsets $U,V$ of $X$ with $X=U\cup V$. Here openness refers to the metric topology of $X$.

It is a path-connected space if, for every $x,y\in X$, there is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$. Such a map is a path. Every path-connected space is a connected space: a separation would pull back along a path joining its two sides to a separation of the interval $[0,1]$.

The arctangent pullback metric is not complete. Take $x_n=n$. Since $\arctan n\to \pi /2$, its transformed sequence is Cauchy, and hence $(x_n)$ is a Cauchy sequence for $\rho$. If it converged in $\rho$ to $x\in \mathbb{R}$, thenwould force $\arctan x=\pi /2$, which is impossible. The missing endpoint of $I$ explains the failure of completeness, even though part (i) shows that the metric topologies agree.

Both properties are preserved by a continuous map.

For sequential compactness, take any sequence $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is sequentially compact, some subsequence satisfies $x_{n_k}\to x\in X$. By continuity, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a sequentially compact space.

For connectedness, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively open sets. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty open sets of $X$ whose union is $X$, contradicting connectedness. Therefore $f(X)$ is a connected space.

The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.