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The uniform convergence condition on a set $X$ isThe index $N$ is independent of $x$; this is stronger than pointwise convergence.

Fix $x_0\in \mathbb{R}$ and $ϵ>0$. Choose one $N$ such that $|f_N(x)-f(x)|<ϵ/3$ for every $x$. The continuity of $f_N$ at $x_0$ gives $\delta >0$ with $|f_N(x)-f_N(x_0)|<ϵ/3$ whenever $|x-x_0|<\delta$. The triangle inequality then givesSince $x_0$ was arbitrary, $\boxed{f\text{ is continuous on }\mathbb{R}}$, the uniform limit theorem. Continuity of each approximant need not be uniform continuity.

A unique factorization domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements, uniquely up to reordering and multiplication of factors by a unit in a ring. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Gauss lemma for polynomials says that over a unique factorization domain $D$, the product of two primitive polynomials is primitive, and the polynomial content satisfies $c(fg)=c(f)c(g)$ up to a unit. In particular a primitive polynomial of positive degree is irreducible over $D$ exactly when it is irreducible over the fraction field of $D$. Consequently a polynomial ring over a unique factorization domain is itself a unique factorization domain.

The Eisenstein criterion says that if $f(X)=a_n{X}^n+\cdots +a_0\in D[X]$ has $n\ge 1$ and there is a prime element $p$ with $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$, then $f$ is irreducible over the fraction field. If $f$ is primitive, it is also irreducible in $D[X]$. Over $\mathbb{Z}$ one may take $p$ to be an ordinary prime number. The primitivity qualification prevents a nonunit constant content from being overlooked.

Every reachable state $i$ reaches zero deterministically in $i$ steps. For $j\in E$, choose a supported lifetime $k>j$; zero can jump to $k-1$ and then count down to $j$. Thus the chain is an irreducible Markov chain on $E$. Its positive-probability first return lengths are exactly the support of $Y_1$; all other return lengths are sums of such lengths. Their greatest common divisor is therefore one by assumption, so the chain is an aperiodic Markov chain.

The equilibrium residual-life distribution isIt is a probability distribution because the tail-sum identity gives ${\sum }_{j\ge 0}\mathbb{P}(Y_1>j)=\mathbb{E}{Y}_1=\mu$. With zero masses outside $E$,so it is a stationary distribution. This also displays why finite mean is needed for normalization.

Use the countable-state Markov chain convergence theorem: an irreducible, aperiodic, positive recurrent discrete-time chain on a countable state space has a unique stationary distribution and $P^n(i,j)\to {\pi }_j$ for every pair of states. All hypotheses were verified, including positive recurrence in part (b). Since $X_0=0$ and $\mathbb{P}(Y_1>0)=1$,The finite-state convergence theorem alone would not suffice when the lifetimes are unbounded. The gcd hypothesis is also necessary for this ordinary limit: deterministic lifetime 2, for example, produces alternating renewal probabilities rather than convergence.

Starting at zero, the next renewal is at time $Y_1$, and there are no intervening renewals because $Y_1\ge 1$. Thus the first return time obeys $T_0=Y_1$ in distribution, indeed under the original renewal construction. Its mean recurrence time isThe chain returns to zero almost surely and has a finite expected return time, so zero is a positive recurrent state. The assumption $\mu <\infty$ is what upgrades recurrence to positive recurrence.

Let $S_k={\sum }_{\ell =1}^k{Y}_{\ell }$, $S_0=0$, be the epochs of the renewal process. The definition uses the next epoch at or after $n$, so $X_n=0$ at an epoch. If $X_n=i\ge 1$, no renewal occurs before $n+i$ and necessarily $X_{n+1}=i-1$.

If $X_n=0$, time $n$ is a renewal epoch, and the next unused interarrival time is independent of the entire observed past with the original distribution. To justify the random index, condition on each possible number $k$ of renewals up to $n$: the event identifying $k$ depends only on $Y_1,\dots ,Y_k$, whereas $Y_{k+1}$ is independent of those variables. Summing over $k$ preserves that distribution and independence. Therefore the residual lifetime Markov chain has transition probabilitiesIts reachable state space is $E=\{j\ge 0:\mathbb{P}(Y_1>j)>0\}$, either an initial finite segment or all nonnegative integers. The conditional next-step law depends only on $X_n$, proving the Markov property. Initially $X_0=0$. The reset is $Y-1$, rather than $Y$, because one unit of time has elapsed by the next step.

For the quantum harmonic oscillator, the time-independent Schrödinger equation iswith $m,\omega ,\hslash >0$. The given dimensionless coordinate yields ${\psi }_{\xi \xi }+(2E/(\hslash \omega )-{\xi }^2)\psi =0$. Substituting the Gaussian factor givesBy the supplied normalizability fact, $f$ is a nonzero polynomial. If its degree is $n$, comparison of the leading coefficient forces $2E/(\hslash \omega )-1=2n$. Conversely, at this value the coefficient recurrenceproduces a degree-$n$ polynomial with parity $n$: start with a nonzero constant for even $n$ or a nonzero linear coefficient for odd $n$, and the recurrence terminates at degree $n$. This is a multiple of the Hermite polynomial. Its product with the Gaussian is square integrable and can be normalized. Therefore all and only the allowed energies are

The interior of the diamond is $B=\{(x_1,x_2):|x_1|+|x_2|<1\}$. Its Minkowski functional, or the uniqueness argument already proved, gives the L1 norm:The final unheaded set $C$ cannot be an open unit ball of a norm, because every norm ball is convex by homogeneity and the triangle inequality. Both $p=(9/10,0)$ and $q=(0,9/10)$ lie in $C$: their squared-distance expression is $101/100>1$. Their midpoint has that expressionso $(p+q)/2\notin C$. Thus $\boxed{\text{no norm has }C\text{ as its open unit ball}}$. The strict inequalities in the definition are respected by this explicit counterexample.

The unheaded preliminary requests are addressed before the first example. A norm is a function $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly for $x=0$, $p(ax)=|a|p(x)$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d(x,y)=p(x-y)$. Translation $T_v$ preserves this distance, and scalar multiplication satisfies $d(ax,ay)=|a|d(x,y)$, so both are continuous; multiplication by zero is constant.

The open unit ball determines the norm uniquely through its Minkowski functional:because $x/t\in B$ is equivalent to $p(x)<t$. This also works for $x=0$.

Under the given radial and convexity hypotheses, set $p_B(0)=0$ and $p_B(v)=1/\lambda (v)$ for $v\ne 0$. The finite symmetric segment along each line gives positivity, absolute homogeneity and $v\in B⟺p_B(v)<1$. If $s>p_B(x)$ and $t>p_B(y)$, then $x/s,y/t\in B$, and the stated convexity condition yields $(x+y)/(s+t)\in B$. Hence $p_B(x+y)<s+t$; taking $s\downarrow p_B(x)$ and $t\downarrow p_B(y)$ proves the triangle inequality. Thus this is a norm determined by a convex radial unit ball. The case $V=\{0\}$ is immediate.

For the coordinate box, the Minkowski functional is the supremum norm:Indeed $x/t\in B$ exactly when $t>{\max }_i|x_i|$.

A nilpotent element belongs to every prime ideal: if $a^r=0\in P$, repeated primality forces $a\in P$. Therefore $N\subseteq {\bigcap }_{P}P$.

Conversely, if $a\notin N$, none of its positive powers is zero. The multiplicative subset $S=\{1,a,a^2,\dots \}$ is disjoint from $(0)$. Part (b) supplies a prime ideal disjoint from $S$, hence not containing $a$. Thus $a$ is outside the intersection, proving the nilradical as the intersection of prime ideals identity:In the zero ring there are no prime ideals; the empty intersection is interpreted as the whole ring, which is also its nilradical. For nonzero rings this argument proves existence of a prime ideal whenever a nonnilpotent element is available.

Consider all ideals containing $I$ and disjoint from the nonempty multiplicative subset $S$. This collection is nonempty because it contains $I$. The ascending chain condition for the Noetherian ring gives a maximal member $P$: otherwise repeatedly choosing a strictly larger member would give an infinite strictly ascending chain of ideals. Since $S$ is nonempty and $P$ misses it, $P\ne R$.

Suppose $ab\in P$ but $a,b\notin P$. Maximality implies that $P+(a)$ and $P+(b)$ both meet $S$, so write $s=p+ra\in S$ and $t=q+ub\in S$ with $p,q\in P$. Thenwhile closure under products gives $st\in S$, a contradiction. HenceThis proves the prime ideal avoiding a multiplicative subset result without needing $1\in S$. The assumptions themselves rule out $0\in S$, since every ideal contains zero.

For nilpotent elements $a,b$ of the commutative ring, choose positive integers $r,s$ with $a^r=b^s=0$. The binomial theorem giveseach term has either $j\ge r$ or $r+s-1-j\ge s$. Also $(-a{)}^r=0$, $0\in N$, and $(ta{)}^r=t^r{a}^r=0$ for every $t\in R$. Thus $N$ is an additive subgroup stable under multiplication by arbitrary ring elements:Commutativity is essential to the simple binomial argument; an arbitrary sum of nilpotent elements in a noncommutative ring need not be nilpotent.

False. Take $\alpha (x,y)=x$, $X=\mathrm{span}\{(1,0)\}$, and $Y=\{0\}$. Then $V=\alpha (X)\oplus \alpha (Y)=\mathbb{R}\oplus \{0\}$, but $X+Y\ne U$ because $(0,1)$ is missing. Thus the premise does not force spanning. It need not force zero intersection either: with the same $\alpha$, take $X=U$ and $Y=\ker \alpha$. Their images still have a direct sum, while $X\cap Y=Y\ne \{0\}$.

False. Take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $\alpha (x,y)=x+y$, $X=\mathrm{span}\{(1,0)\}$, and $Y=\mathrm{span}\{(0,1)\}$. Then $U=X\oplus Y$ is a direct sum, but $\alpha (X)=\alpha (Y)=\mathbb{R}$, so their intersection is not zero. Their ordinary sum is $V$, as always follows from surjective function and $U=X+Y$; the failure is precisely the direct sum assertion.

True. Use the lifted basis $u_1,\dots ,u_r$ above and put $W=\mathrm{span}\{u_1,\dots ,u_r\}$. If $\sum c_i{u}_i=0$, applying $\alpha$ gives $\sum c_i{v}_i=0$, so all $c_i=0$. Thus the $u_i$ form a basis of $W$, and $\alpha {|}_W$ takes this basis to the basis of $V$. It is a linear isomorphism.

In fact the right inverse gives the stronger splittingFor every $u\in U$, write $u=(u-\beta \alpha u)+\beta \alpha u$; the first term is in the kernel and the second in $W$. Their intersection is zero because $\alpha {|}_W$ is injective. This is a direct sum decomposition.

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.

False. A left inverse would force $\alpha$ to be injective: if $\alpha u=0$, then $u=\beta \alpha u=0$. Take the surjective linear map $\alpha :{\mathbb{R}}^2\to \mathbb{R}$ given by $\alpha (x,y)=x$. Since $(0,1)$ belongs to its kernel, every composite $\beta \alpha$ annihilates $(0,1)$ and cannot be the identity map on ${\mathbb{R}}^2$.