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If $A$ is closed and $a_n\in A$ converges to $x\in X$, then $x$ cannot lie in the open complement: an open ball about such an $x$ would eventually contain all $a_n$ while missing $A$. Thus $x\in A$.

Conversely, suppose $A$ contains every limit of a sequence in $A$. If $x\in \overline{A}$, every ball $B(x,1/n)$ meets $A$; choose $a_n$ in that intersection. Then $a_n\to x$, so $x\in A$. Hence $\overline{A}=A$, proving that a subset of a metric space is a closed set exactly when it is sequentially closed.

Show that the subspace $C$ of convergent real sequences is closed in $X$. Suppose $a^{(m)}\in C$ converges uniformly in the coordinate index to $a\in X$. Given $ϵ>0$, choose $m$ so ${\sup }_j|a_j-a_j^{(m)}|<ϵ/3$. Since the fixed sequence $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $ϵ/3$. ThenThus $a$ is a Cauchy sequence of real numbers and converges by completeness of $\mathbb{R}$. Hence $a\in C$, so $C$ is closed. Part (a)(iii) proves the convergent-sequence subspace is complete.

Let $a^{(m)}=(1,1/2,\dots ,1/m,0,0,\dots )$. Each has finite support, and in the supremum norm metric it converges to $a=(1,1/2,1/3,\dots )$, because $d(a^{(m)},a)=1/(m+1)$.

The limit is bounded but does not have finite support. Thus the finite-support subspace is not closed in $X$, and by part (a)(iii) it is not complete. Equivalently, the displayed sequence is Cauchy in that subspace but has no limit there.

Write a first return time as $2J$, where $\mathbb{P}(J=j)=2^{-j}$ for $j\ge 1$. It has mean $2\cdot 2=4$ and period two, so the stationary probability at one is $1/4$. The periodic convergence theorem would give limits $1/2$ along even times and zero along odd times.

Here one can obtain them directly. Put $u_n={\mathbb{P}}_1(X_{2n}=1)$, with $u_0=1$. Splitting according to the first return gives the renewal equationInduction gives $u_n=1/2$ for every $n\ge 1$: the right side is $2^{-n}+\frac{1}{2}{\sum }_{j=1}^{n-1}{2}^{-j}=1/2$. Odd-time returns are impossible because every return interval is even. Therefore the exact set of limit points isThe sequence does not converge to the stationary mass $1/4$, precisely because it is periodic.

A QR decomposition is $\boxed{A=QR}$ with $Q$ orthogonal and $R$ upper triangular. One algorithm for computing it is Householder reflection reduction; a sequence of such reflections produces a QR factorization. The diagonal of $R$ may include zero when $A$ is singular, and no nonsingularity assumption is needed here.

For each $n$, let $N_n$ be the null set where $f_n\ne g_n$. Their union $N$ is null. At any $x\notin N$, the two sequences agree term by term, so their pointwise limits are equal: $f(x)=g(x)$. Therefore $f\sim g$. Taking the countable union first is essential: it produces one exceptional set outside which all equalities hold simultaneously.

No such sequence exists. By Fatou's lemma on $(0,1)$,But the proposed nonnegative measurable integrand is positive at every point of $(0,1)$, so its integral cannot be zero. Explicitly, the sets where it is at least $1/k$ cover $(0,1)$; at least one has positive measure, forcing a positive integral. This contradicts the displayed bound.

For each integer $m\ge 1$, list the $m^2$ functionsin increasing $j$, and concatenate the blocks in increasing $m$. Each function has $L^1$ norm $1/m$, so the resulting sequence has norm tending to zero. Every $x\in (0,1)$ lies in exactly one interval of each block, where the function takes value $m$. ThereforeThis unbounded typewriter sequence with vanishing L1 norm is a moving-spike construction: the support shrinks fast enough to beat the increasing height.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence

A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.

If $A$ is closed and $a_n\in A$ converges to $x\in X$, then $x$ cannot lie in the open complement: an open ball about such an $x$ would eventually contain all $a_n$ while missing $A$. Thus $x\in A$.

Conversely, suppose $A$ contains every limit of a sequence in $A$. If $x\in \overline{A}$, every ball $B(x,1/n)$ meets $A$; choose $a_n$ in that intersection. Then $a_n\to x$, so $x\in A$. Hence $\overline{A}=A$, proving that a subset of a metric space is a closed set exactly when it is sequentially closed.

Show that the subspace $C$ of convergent real sequences is closed in $X$. Suppose $a^{(m)}\in C$ converges uniformly in the coordinate index to $a\in X$. Given $ϵ>0$, choose $m$ so ${\sup }_j|a_j-a_j^{(m)}|<ϵ/3$. Since the fixed sequence $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $ϵ/3$. ThenThus $a$ is a Cauchy sequence of real numbers and converges by completeness of $\mathbb{R}$. Hence $a\in C$, so $C$ is closed. Part (a)(iii) proves the convergent-sequence subspace is complete.

Let $a^{(m)}=(1,1/2,\dots ,1/m,0,0,\dots )$. Each has finite support, and in the supremum norm metric it converges to $a=(1,1/2,1/3,\dots )$, because $d(a^{(m)},a)=1/(m+1)$.

The limit is bounded but does not have finite support. Thus the finite-support subspace is not closed in $X$, and by part (a)(iii) it is not complete. Equivalently, the displayed sequence is Cauchy in that subspace but has no limit there.

Weak convergence means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. Norm convergence implies this by $|\ell (u_j-u)|\le \Vert \ell \Vert \Vert u_j-u\Vert$. The converse fails for coordinate unit vectors in ${\ell }^2$: their norms are one, but pairing with any square-summable sequence tends to zero, so they are weakly null.

For the subsequence assertion apply Gram-Schmidt process to a countable dense subset to choose a countable orthonormal basis $(e_m)$ of the separable Hilbert space and a bound $M$ for the sequence. Repeated subsequence selection and diagonalization make every coefficient $⟨u_{j_k},e_m⟩$ converge to a scalar $a_m$. Bessel inequality gives ${\sum }_{m=1}^N|a_m{|}^2\le M^2$ for every $N$, so $u={\sum }_m{a}_m{e}_m$ exists in $H$ with norm at most $M$. Pairings with finite linear combinations of the basis converge. Approximating any other test vector by such combinations and using the common norm bound proves $u_{j_k}\rightharpoonup u$. The finite-dimensional case uses the same argument with a finite basis.

Apply this to every subsequence of the bounded $L^2$ sequence. Any weakly convergent further subsequence has limit $v$ with ${\int }_a^{b}v=0$, since interval indicators are $L^2$ tests. The permitted fact gives $v=0$ almost everywhere. If the whole sequence failed to converge weakly to zero, some fixed test functional and subsequence would stay a positive distance from zero. Its weakly convergent further subsequence would then contradict the preceding conclusion. Thus the entire sequence converges weakly to zero in $L^2(0,1)$.

The De la Vallée-Poussin criterion characterizes uniform integrability by existence of a nonnegative increasing convex function $\Psi$ with $\Psi (s)/s\to \infty$ and a uniform bound on $\int \Psi (|v_j|)$. On a finite interval this gives a weakly convergent subsequence in $L^1$ by the Dunford-Pettis theorem.

After discarding a finite prefix of the minimizing sequence if needed, put $H(t)=\Phi (t)-M\ge 0$. Its derivative sequence has $\int H(u_j^{\prime })\le C$. The superlinear bound directly gives uniform integrability: for $H(t)\ge A|t|$ outside $[-R_A,R_A]$,If the convex version of the criterion is used, choose increasing $R_m\to \infty$ so $H(t)\ge m^2|t|$ for $|t|\ge R_m$, and set $\Psi (s)={\sum }_{m\ge 1}(s-R_m{)}_{+}$. This is locally a finite sum, convex and increasing. It is superlinear because for fixed $N$ and $s\ge 2R_N$, $\Psi (s)\ge Ns/2$. If exactly $m$ terms are active, $\Psi (s)\le ms\le H(t)$ for $s=|t|$, so its integrals have the same bound. The criterion therefore supplies $u_{j_k}^{\prime }\rightharpoonup v$ in $L^1$.

Define $u(x)=\alpha +{\int }_a^{x}v$. Interval-indicator tests give $u_{j_k}(x)\to u(x)$ for every $x$, including $b$, so its endpoint values are correct and $u\in W^{1,1}$. The common derivative $L^1$ bound makes these functions uniformly bounded. Uniform integrability also gives common equicontinuity, since an increment is bounded by the derivative integral on the intervening interval. The Arzelà-Ascoli theorem yields uniform convergence to the same identified limit, and in particularHere the printed admissible space is $W^{1,1}$, so this compactness argument gives precisely the needed membership.

Superlinearity and continuity make $\Phi$ bounded below by a finite constant $M$. Consequently $I(u)\ge M(b-a)$ for every admissible function. The affine endpoint interpolant has finite energy by continuity on its compact range, so the infimum is finite. Some other admissible energies can be infinite; this does not affect the bounded-below or minimizing-sequence argument.

The assumed weak-star limit is the constant $m=\theta a+(1-\theta )b$. Each integer rescaling has exactly $j$ full periods on $(0,1)$, so a change of variables on those periods givesApply the asserted lower semicontinuity to this sequence. The integral of $f$ of its constant limit is $f(m)$, givingThis holds for every real $a,b$ and $0<\theta <1$; the endpoint values of $\theta$ give equality. Therefore $f$ is convex.

For every fixed truncation index $N$, the remainder must be smaller than the last retained term:This is an asymptotic expansion relative to the sequence in part (i). The number of retained terms is fixed before taking the limit. The associated series need not converge for any nonzero parameter; its role is to give successively smaller truncation errors. If $a_n$ is an asymptotic sequence, the same condition can be stated as a remainder $O(a_{N+1})$ when the next expansion condition is available, but the displayed little o notation formulation is the defining one.

The tail sigma-algebra of this sequence isFor each fixed $r$, the preceding event equality givesEvery event on the right is determined by coordinates with index at least $r$. Therefore $A\in \sigma (X_r,X_{r+1},\dots )$ for every $r$, and $A\in \mathcal{T}$. The random variables $X_n$ are independent, so Kolmogorov's zero–one law applies: every tail event has probability zero or one. Hence $\mathbb{P}(A)\in \{0,1\}$.

The role of independence can also be seen directly. A tail event is independent of each finite initial sigma-algebra $\sigma (X_1,\dots ,X_m)$. The Monotone class theorem extends this independence to $\sigma (X_1,X_2,\dots )$; since $A$ itself lies there, it is independent of itself. Thus $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.^[ref]

The conditions on the cumulative distribution function imply $0<X_n\le 2$ almost surely. With ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$, adjoin $Y_0=1$ and the trivial ${\mathcal{H}}_0$. The random variables $Y_n$ are nonnegative, adapted, and bounded by $2^n$ at each fixed time. Independence and $\mathbb{E}{X}_{n+1}=1$ giveThus this is the product martingale from independent mean-one factors, with $\mathbb{E}{Y}_n=1$. The almost sure supermartingale convergence theorem states that a nonnegative supermartingale converges almost surely to a finite integrable random variable. Applied to this nonnegative martingale, it shows $Y_n\to Y_{\infty }<\infty$ almost surely. Every convergent real sequence is bounded, so ${\sup }_{n\ge 1}{Y}_n<\infty$ almost surely.

The printed maximum can in fact be taken literally here. If $X_1=1$ almost surely, all products equal $1$. Otherwise the strict Jensen equality criterion, applied to the strictly convex function $t\mapsto -\sqrt{t}$ on $(0,\infty )$, gives $q=\mathbb{E}\sqrt{X_1}<1$. Independence yields $\mathbb{E}\sqrt{Y_n}=q^n$. By Markov's inequality, ${\sum }_n\mathbb{P}(Y_n>\epsilon )\le {\epsilon }^{-1/2}{\sum }_n{q}^n<\infty$ for every $\epsilon >0$. The First Borel-Cantelli lemma and the countable choices $\epsilon =1/m$ give $Y_n\to 0$ almost surely. Since $Y_1>0$, a sequence with this limit attains its positive supremum among finitely many initial terms. This is an instance of fractional moment decay of independent mean-one products.