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For a nondecreasing function, the endpoint bounds show it is bounded. On the uniform partition into $N$ intervals, its Darboux sums satisfyFor a nonincreasing function apply the same calculation to $-f$, or use $|f(1)-f(0)|/N$. Thus every monotone function on $[0,1]$ is Riemann integrable.

For the unheaded rational-weight example, each summand is nonnegative, and increasing $x$ only adds indices. Hence its sum is nondecreasing and lies between zero and one, including the separately specified value at zero. It is a left-continuous cumulative function of an atomic measure.

If $q_j\in [0,1)$ and $q_j<x\le 1$, the index $j$ appears in the sum at $x$ but not at $q_j$. All other changes are nonnegative, so $f(x)-f(q_j)\ge 2^{-j}$. This remains true for arbitrarily small positive $x-q_j$, proving discontinuity at $q_j$, including a right discontinuity at zero when that rational is encountered. Every interval of positive length contains such a rational in its interior. The discontinuities are therefore dense, although the function is Riemann integrable by the monotone-function result just proved.

More concretely, truncate after the first $N$ weights. The resulting finite sum of step functions is integrable, and the uniform tail is at most $2^{-N}$. This also proves integrability and gives, if its value is desired,The integral formula follows by integrating the finite sums and using the uniform tail bound. Here “every interval” has the usual nondegenerate-interval meaning; a singleton is not being asserted to contain a discontinuity.

For a bounded real function on $[0,1]$, take a partition $P$ with nodes $0=x_0<\cdots <x_N=1$. Its lower Darboux sum and upper Darboux sum areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; their common value is its Riemann integral. Equivalently, for every $\eta >0$ some partition has $U-L<\eta$. Indeed a common refinement of nearly extremizing upper and lower partitions proves necessity, while arbitrarily small gaps force equality of the two integrals. This is the Riemann integrability criterion.

A continuous function on a compact interval is bounded and uniformly continuous. Choose a mesh small enough that its oscillation on each subinterval is less than $\eta$. Then $U-L\le \eta {\sum }_j(x_j-x_{j-1})=\eta$; using $\eta /2$ first makes the desired inequality strict. Thus every continuous function on $[0,1]$ is Riemann integrable.

From $f^{\prime \prime }=f$, differentiating successively gives $f^{(2r)}=f$ and $f^{(2r+1)}=f^{\prime }$ whenever those derivatives have been constructed. Starting with twice differentiability, this relation supplies the next two derivatives at each stage. Thus $f$ is infinitely differentiable, and its origin series isFor a fixed $x$, both $f$ and $f^{\prime }$ are bounded on the compact interval between zero and $x$, say by $M_x$. Every higher derivative is one of these two functions. Taylor's theorem, with the remainder just proved, therefore bounds the order-$n$ remainder by $M_x|x{|}^n/n!$, which tends to zero: its consecutive positive-term ratio is $|x|/(n+1)$. Hence the series converges to $f(x)$ for every real $x$. It also gives $f(x)=A\cosh x+B\sinh x$ by the usual hyperbolic-function series.

At any other centre $a$, apply the proved origin result to $g(t)=f(a+t)$, which also satisfies $g^{\prime \prime }=g$. Its even derivatives at zero are $f(a)$, and its odd ones are $f^{\prime }(a)$. ConsequentlyThis proves convergence at every centre without assuming that smoothness alone implies equality to a Taylor series.

For $x\ne 0$, put $P(t)={\sum }_{k=0}^{n-1}{f}^{(k)}(0)t^k/k!$ and $K=(f(x)-P(x))/x^n$. The auxiliary function $F(t)=f(t)-P(t)-K{t}^n$ satisfiesRolle's theorem first gives a zero of $F^{\prime }$ strictly between zero and $x$. Apply Rolle's theorem to $F^{\prime }$ between that zero and zero, where $F^{\prime }(0)=0$, to obtain a zero of $F^{\prime \prime }$. Continue in this way. The derivatives through order $n-1$ are continuous, since they are differentiable; hence every application is valid even if $f^{(n)}$ is not continuous. After $n$ applications there is $c$ strictly between zero and $x$ with $F^{(n)}(c)=0$.

But $F^{(n)}(c)=f^{(n)}(c)-n!K$, so $K=f^{(n)}(c)/n!$. Writing $c=\theta x$ yields the Taylor formula with Lagrange remainderThe same construction using repeated applications of Rolle's theorem applies when $x<0$, with intervals ordered from $x$ to zero. For $x=0$ the identity is immediate and any $\theta \in (0,1)$ works. This proves the Taylor theorem with Lagrange remainder using only the allowed theorem and elementary differentiation.

Let $D=f_{+}^{\prime }(a)$ and write $g(t)=(f(t)-f(a))/(t-a)$ for $t>a$. By the assumed right derivative, choose $\delta >0$ with $|g(t)-D|<ϵ/2$ whenever $a<t<a+\delta$. Choose $b_0\in (a,\min (b,a+\delta ))$.

Apply the mean value theorem to $f$ on $[a,b_0]$. It gives $x\in (a,b_0)$ with $f^{\prime }(x)=g(b_0)$. Both $g(x)$ and $g(b_0)$ are within $ϵ/2$ of $D$, soThis proves the endpoint secant-tangent approximation without any continuity assumption on $f^{\prime }$. The continuity required by the theorem is only that of $f$ itself.

For continuity at any $x$, choose $\delta ={\eta }^{1/\alpha }$ for a desired error $\eta >0$. If $|y-x|<\delta$, the stated bound gives $|f(y)-f(x)|<\eta$. Thus $f$ is continuous, in fact uniformly continuous.

When $\alpha >1$, the difference quotient satisfiesTherefore the derivative exists at every point and is zero. Part (i) proves $f$ is constant. This is the lemma that Hölder exponent greater than one forces constancy; no differentiability was assumed in advance.

The mean value theorem states that a function continuous on $[p,q]$ and differentiable on $(p,q)$ has some $c\in (p,q)$ with $f(q)-f(p)=f^{\prime }(c)(q-p)$.

Take any $p<q$ inside the given open interval. Differentiability implies the continuity needed on $[p,q]$, and the derivative is zero throughout it. The mean value theorem therefore gives $f(q)=f(p)$. Since the two points were arbitrary, the function is constant on the interval.

Let $C$ and $D$ be the connected components of $x$ and $y$. First, $C\times D$ is connected in the product topology. For each $c\in C$, the connected slice $\{c\}\times D$ meets the connected slice $C\times \{y\}$. Their union over $c$ is connected by the common-intersection argument, and is exactly $C\times D$.

Now let $K$ be the connected component of $(x,y)$ in $X\times Y$. The coordinate projections are continuous, so their images of $K$ are connected and contain $x$ and $y$. Maximality therefore gives $K\subseteq C\times D$. Conversely, the connected set $C\times D$ contains $(x,y)$, so it lies in $K$. $K=C\times D$. This is connected components of a product space; it requires neither local connectedness nor path connectedness.

Fix $x$, and let $C_x$ be the union of all connected subsets containing $x$. Their union is connected: a separation of the union would place $x$ on one side, and every connected member meeting that side would have to lie entirely there. The other side would be empty. By the defining equivalence relation this union is exactly the connected component of $x$.

The closure of a connected set is connected as well. Indeed, if its closure had a separation into relatively open nonempty pieces, density would make both pieces meet the original set and give a separation of that connected set. Therefore $\overline{C_x}$ is a connected subset containing $x$, so maximality gives $\overline{C_x}\subseteq C_x$. The reverse inclusion is automatic. Every connected component is connected and closed.

The map $\phi :M\to N$, $x\mapsto rx$, is a module homomorphism, and it is injective because $r\ne 0$ and $M$ is a torsion-free module. The linear independence in a module of $T$ makes the map $R^{|T|}\to N$ sending the standard basis to $T$ an isomorphism. Thus $N$ is a finite free module, and $M\cong \phi (M)\subseteq N\cong R^{|T|}$. This proves embedding a finitely generated torsion-free module in a finite free module. No claim that the submodule itself is free over an arbitrary integral domain is needed.

For each $s\in S\setminus T$, maximality implies a nontrivial relationHere $a_s\ne 0$, since otherwise this would contradict the linear independence in a module of $T$. Thus $a_{s}s\in N$. Take $r={\prod }_{s\in S\setminus T}{a}_s$, with empty product equal to one. Because $R$ is an integral domain, $r\ne 0$. Every $s\in S\setminus T$ satisfies $rs\in N$, and this is also true for $s\in T$. Expressing any element of the finitely generated module as an $R$-linear combination of $S$ now gives $rM\subseteq N$. This is clearing denominators relative to an independent module subset; it does not require $R$ to be a field.

False without a separation assumption. Give $X=\{0,1\}$ the indiscrete topology $\{\varnothing ,X\}$, and take $Y=\{0\}$. Both spaces are finite and hence compact, but $Y$ is not a closed set because $X\setminus Y=\{1\}$ is not open. In a Hausdorff space, a compact subset would be closed; that hypothesis is absent here.

True. In a discrete topology, the singletons form an open cover. If $X$ is compact, a finite subcover exists, so $X$ is finite. Conversely, any open cover of a finite set has a finite subcover: choose one member containing each point. This converse does not require the discrete topology.

For an active monster, independence givesBoth probabilities vanish before time $100$ and at odd times. By convolution and Cauchy-Schwarz, $p_{2m}(a,0)\le p_{2m}(0,0)<1/m$: indeed, write the $2m$-step probability as the inner product of two translates of the $m$-step probability mass function. ThereforePositivity follows by prescribing $100$ simultaneous steps taking both walkers straight to the origin; that event has probability $4^{-200}$. An active monster thus leaves a positive escape probability.

For a sleepy monster, the origin capture by a sufficiently sleepy planar random walk can be proved using the printed schedule: during $I_j=\{n_j+1,\dots ,n_{j+1}\}$ it occupies $Y_{j+1}$. For any $M$, let $J_M$ be the first $j\ge M$ with $Y_{j+1}=0$. Markov-chain recurrence of $Y$ makes $J_M$ finite almost surely, and it is independent of $X$. By conditioning on $J_M$ and using the stated block bound,As $M\to \infty$, continuity of probability for decreasing events shows that the event $C_{\infty }$ of infinitely many simultaneous origin visits has probability at least $1/2$. Regard the independent trajectories as continuing even after a possible first capture.

To turn this into probability one, use the eventual coupling of planar symmetric random walks; the block events need not be independent. Two planar simple symmetric random walks begun at sites of the same parity can be coupled to agree eventually. Run them independently until they meet. Their difference walk is irreducible on the even-parity sublattice, and its $m$-step return probability is $p_{2m}(0,0)$, by convolution and symmetry. The preceding divergent-series criterion makes this difference walk a recurrent Markov chain, so it hits zero almost surely. After that stopping time, use identical increments for both copies, preserving their marginal laws by the Strong Markov property.

Fix any finite calendar time $N$ and any two possible histories up to that time. Corresponding princess positions have the same parity, as do the corresponding monster positions at its internal walk time. Apply the eventual probabilistic coupling separately to the two princess futures and the two monster futures, keeping the princess and monster walks independent in each pair. Both the calendar clock and the monster's internal clock tend to infinity. Hence after a finite calendar time the two pairs of trajectories agree, so either both realize $C_{\infty }$ or neither does. The conditional probability of $C_{\infty }$ is therefore the same for every possible finite history.

Consequently $\mathbb{E}(1_{C_{\infty }}\mid {\mathcal{F}}_N)=\mathbb{P}(C_{\infty })$ for every $N$. By the martingale convergence theorem, these conditional expectations converge to $1_{C_{\infty }}$ as the histories exhaust the trajectories. Thus $\mathbb{P}(C_{\infty })$ is zero or one. Since it is at least $1/2$, it is one. In particular,This argument uses the supplied unconditional block probabilities; it does not silently strengthen them to conditional probabilities given past failures.

For a planar simple symmetric random walk write its coordinates as $(S_n,T_n)$ and rotate to $(S_n+T_n,S_n-T_n)$. Each step now has four equally likely values $(1,1),(1,-1),(-1,1),(-1,-1)$. Thus the two rotated coordinates are independent one-dimensional simple symmetric random walks.

A return is impossible at odd times, and at time $2n$ its probability isFor $n\ge 2$, the supplied estimate gives $1/(4n)<p_{2n}(0,0)<1/n$. At $n=1$ the lower bound is actually equality, $p_2(0,0)=1/4$; this harmless strict-endpoint typo does not affect the argument. The harmonic series diverges, so the criterion proved above shows that the origin is recurrent, in the terminology of a recurrent Markov chain. Translation invariance gives the same conclusion for every state. The Markov chain is irreducible, so from any starting point it hits any fixed state almost surely and revisits it infinitely often. One can see the hitting assertion directly: repeated returns to the starting point give repeated positive-probability opportunities to follow a finite path to the target.

Let $T_j=\inf \{n\ge 1:X_n=j\}$. By the Strong Markov property, the probability of hitting $j$ and subsequently returning to $i$ is $f_{ij}{f}_{ji}$. This event implies a return to $i$, so $f_{ii}\ge f_{ij}{f}_{ji}$. For $i=j$, interpret the event as two successive returns.

Set $T_i^{(0)}=0$ and let $T_i^{(r)}$ be the $r$th successive return. Repeated use of the Strong Markov property gives ${\mathbb{P}}_i(T_i^{(r)}<\infty )=f_{ii}^r$. The total number $N_i$ of visits, including time zero, is both ${\sum }_{n\ge 0}{1}_{\{X_n=i\}}$ and ${\sum }_{r\ge 0}{1}_{\{T_i^{(r)}<\infty \}}$. By Tonelli theorem,The identity is valid with value $+\infty$. In particular, divergence of the return-probability sum is equivalent to Markov-chain recurrence.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrödinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

The time-independent Schrödinger equation gives the even interior wavefunction $\psi =A\cos (\alpha x)$, with ${\alpha }^2=2m(U+E)/{\hslash }^2$. Outside, normalizability of a bound state requires $\psi =C{e}^{-\beta (|x|-l)}$, with ${\beta }^2=-2mE/{\hslash }^2$.

Continuity of the wavefunction and its derivative at $x=l$ gives $C=A\cos (\alpha l)$ and $-\beta C=-\alpha A\sin (\alpha l)$. A nonzero solution cannot have $\cos (\alpha l)=0$, so division givesAt the formally allowed endpoint $E=-U$, the even interior solution is constant; derivative matching would force the exterior amplitude to vanish, leaving the zero solution. Nontrivial bound states therefore have $-U<E<0$. The matching formula is the even-state condition for a finite square well.

For $z=R+iy$, $0\le y\le \pi$, the numerator modulus is $e^{-2Ry/\pi }$ and the denominator modulus is at least $1-e^{-2R}$. Hence the magnitude of the right-side contour integral is at mostOn $z=-R+iy$ the corresponding bounds are $e^{2Ry/\pi }$ and $e^{2R}-1$, giving exactly the same bound $\pi /(2R)$. Thus the total vertical contribution is $O(R^{-1})$ and vanishes. This integral estimate handles the endpoints where a pointwise uniform-decay argument would fail.

On the top side put $z=x+i\pi$, with $x$ running from $R$ to $-R$. Since $e^{i(x+i\pi {)}^2/\pi }=-e^{i{x}^2/\pi }{e}^{-2x}$ and $e^{-2(x+i\pi )}=e^{-2x}$, the reversed orientation cancels the minus sign. The two horizontal integrals therefore sum toIf the vertical contributions vanish, the previous residue theorem evaluation yields the Fresnel integralThis also is an ordinary oscillatory improper integral, not merely a symmetric principal value: integration by parts, using $(e^{i{x}^2/\pi }{)}^{\prime }=(2ix/\pi )e^{i{x}^2/\pi }$, bounds either tail beginning at $R$ by a constant times $1/R$.