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Replace the two $G$ quantiles by conditional bootstrap quantiles ${\hat{q}}_u$ in the interval from (a). Uniform convergence implies ${\hat{q}}_u\to q_u$ in probability at continuity points, so Slutsky and continuity of $G$ give coverage tending to $1-\alpha$.

Take $0\le r<R$. At the point $r$, the power series has absolute convergence, so ${\sum }_{n=0}^{\infty }|a_n|r^n<\infty$. On the whole interval $[-r,r]$,The Weierstrass M-test with $M_n=|a_n|r^n$ proves uniform convergence on $[-r,r]$. It also gives the explicit tail estimateFor $r=0$ the interval consists only of zero, and all positive-degree terms vanish. If $R=\infty$, the same argument applies to every finite $r$; if $R=0$, there is no $r\ge 0$ with $r<R$.

The Weierstrass M-test states: if functions $f_n:E\to \mathbb{R}$ or $\mathbb{C}$ satisfythen $\sum f_n$ has uniform convergence and absolute convergence on $E$. In fact $\sum |f_n|$ also has uniform convergence, since every tail is bounded by the corresponding tail of $\sum M_n$:

Write the partial sums as $S_N(x)={\sum }_{n=1}^N{f}_n(x)$. The series has uniform convergence to $s:E\to \mathbb{R}$ ifThe same threshold $N_0$ must control every point of $E$. Equivalently, the supremum error ${\sup }_{x\in E}|S_N(x)-s(x)|$ tends to zero, for nonempty $E$.

For real-valued functions, the Uniform Cauchy criterion applied to partial sums says that $\sum u_n$ has uniform convergence exactly whenSufficiency uses completeness of the real numbers.

Let $M$ be a common bound for the uniformly bounded family of functions $(g_n)$; increase it if necessary so that $M>0$. The assumed uniform absolute convergence of $\sum f_n$ gives $N$ such thatThe triangle inequality now yieldsThe Uniform Cauchy criterion therefore proves uniform convergence of $\sum f_n{g}_n$. The same estimate with $|f_n{g}_n|$ proves uniform absolute convergence as well.

Choose the unbounded function $\boxed{g(x)=1/x}$ on $E=(0,1/2]$. It is unbounded because $g(1/k)=k$ for all integers $k\ge 2$. For each $x\in E$, the geometric series gives the sum and its remainder after degree $N$:Since $0<x\le 1/2$,This proves uniform convergence. Although the first term $1/x$ and the sum are unbounded, the late tails have a bound tending to zero independently of $x$. Uniform convergence concerns those tails and does not require the individual functions to be bounded.

Let $S_N={\sum }_{n=1}^N{f}_n$ and let $s$ be its uniform limit. Choose a finite $B\ge 0$ with $|g(x)|\le B$ for all $x\in E$. The partial sums of the multiplied series are $g{S}_N$, andFor any $\epsilon >0$, uniform convergence supplies $N_0$ making $|S_N(x)-s(x)|<\epsilon /(B+1)$ for all $N\ge N_0$ and $x\in E$. The displayed estimate then gives an error less than $\epsilon$. Thus

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

The sequence $(f_n)$ converges by uniform convergence to $f$ on $E$ ifThe index $M$ must work simultaneously for every $x\in E$.

Given $\epsilon >0$, choose $m\ge N$ so large that $|f_m(x)-f(x)|<\epsilon /3$ for every $x\in E$. The continuity of this one fixed $f_m$ at $x_0$ gives $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_m(x)-f_m(x_0)|<\epsilon /3$. The triangle inequality then givesThus $f$ is continuous at $x_0$, with continuity understood relative to $E$. This is the pointwise form of the uniform limit theorem.

No. Let $E=\mathbb{R}$, $x_0=0$ andThe supremum norm satisfies $\Vert f_n-f{\Vert }_{\infty }=1/n\to 0$, so there is uniform convergence to the continuous function $f$. But every $f_n$ is discontinuous at zero: $f_n(0)=1/n$ whereas $f_n(x)=0$ for every $x\ne 0$. No tail consists of functions continuous at $x_0$.

The finite geometric series identity gives, for each $|x|<1$,For every fixed $n$, the absolute value of this error tends to infinity as $x\to 1^{-}$. Thusand there is no uniform convergence on $(-1,1)$.

At $x=0$, $f_n(x)=0$. For fixed $x>0$,Thus the pointwise limit is zero everywhere on $[0,1]$. However, with $t=nx$, the inequality $1+t^2\ge 2t$ gives $f_n(x)\le 1/2$, and equality occurs at $x=1/n\in [0,1]$. Henceso there is no uniform convergence. The moving evaluation point $1/n$ captures a peak missed by each fixed positive input.

Yes. By uniform convergence, choose $N$ so that $|p_n(x)-f(x)|<1$ for every $n\ge N$ and every $x\in \mathbb{R}$. ThenThe polynomial $p_n-p_N$ is bounded on the whole real line. The supplied growth property of nonconstant polynomials implies that this difference must be constant; write $p_n=p_N+c_n$. Evaluating at zero gives $c_n=p_n(0)-p_N(0)\to c=f(0)-p_N(0)$. For each $x$, take the limit of a sequence in $p_n(x)=p_N(x)+c_n$ to obtainThis is a polynomial. The uniform limits of polynomials on the real line therefore have a tail in which all nonconstant coefficients have already stabilized; only the constant coefficients may continue to change.

Put $E_n={\sup }_{t\in [a,b]}|f(t)-{\varphi }_n(t)|\to 0$, so the hypothesis is uniform convergence. Every step function is bounded and Riemann integrable: away from its finitely many partition points it is constant, and those exceptional points can be covered by intervals of arbitrarily small total length. A partition of an interval containing their boundaries makes the difference of its upper Darboux sum and lower Darboux sum arbitrarily small.

Choose one $n$ with $E_n<\infty$. Boundedness of ${\varphi }_n$ and $|f|\le |{\varphi }_n|+E_n$ show that $f$ is a bounded function. Assume $a<b$, since for $a=b$ all the integrals are zero. Given $\eta >0$, choose $n$ so that $2E_n(b-a)<\eta /2$, and then a partition of an interval $P$ such that $U({\varphi }_n,P)-L({\varphi }_n,P)<\eta /2$. On every partition subinterval, $|f-{\varphi }_n|\le E_n$ givesThusThe Riemann integrability criterion proves that $f$ is Riemann integrable. Finally, linearity of the integral and monotonicity of the Riemann integral giveThereforeThis proves the relevant case of uniform convergence preserves integrals on a compact interval.

The printed assertion needs an endpoint qualification: the enumeration includes zero, but $1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed discontinuities occur at every $r_j>0$.

Define the finite step function sumsThe assumed convergence of ${\sum }_{k}1/k^2$ and the bound on each indicator function giveThus the defining series converges everywhere and has uniform convergence. Part (b) proves that $f$ is Riemann integrable; it also gives, if desired,For $r_j>0$ and $0\le t<r_j$, every indicator function is nonincreasing, and the $j$th drops by one between $t$ and $r_j$. HenceTaking inputs from the left tending to $r_j$ proves a discontinuity there.

For completeness, continuity at zero follows directly from a finite-head and tail argument. Given $\epsilon >0$, choose $N$ with ${\sum }_{k>N}1/k^2<\epsilon$. Choose $\delta >0$ below every positive $r_k$ with $k\le N$. For $0\le t<\delta$, the first $N$ indicator functions have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\le {\sum }_{k>N}1/k^2<\epsilon$. This confirms that the endpoint exception is real. Replacing the printed enumeration of $[0,1)$ by one of $(0,1)$ would remove it.

Finally, the density of the rational numbers supplies a positive rational $r_j$ in every nonempty open subinterval of $[0,1]$. Thus a Riemann integrable function can have a discontinuity in every open subinterval: dense jumps do not prevent Riemann integrability.

Subtract the constant steady state profile: set $v=T-T_0$. ThenThe last equation follows from Fourier's law and zero heat flux at the insulated end. For a nonzero separated solution $v=X(x)H(t)$, separation of variables givesTo see that $\mu >0$ for a nonzero eigenfunction, multiply the spatial equation by $X$ and use integration by parts:Both boundary terms vanish. If the right side were zero, $X$ would be constant and $X(0)=0$ would make it zero. Thus write $\mu ={\lambda }^2$ with $\lambda >0$. Solving gives $X=A\sin (\lambda x)$, and $X^{\prime }(L)=0$ requires $\cos (\lambda L)=0$. The mixed Dirichlet-Neumann modes on an interval therefore haveThe eigenvalues of $-d^2/d{x}^2$ are ${\mu }_n={\lambda }_n^2$; ${\lambda }_n$ are their positive square roots. By superposition, with coefficients for which the resulting series and its required derivatives converge at positive times,The eigenfunctions are orthogonal on $[0,L]$ and have squared norm $L/2$. Expanding the initial excess temperature $v(x,0)=T_{0}x/L$ givesThus the requested temperature isAt the insulated endpoint $\sin ({\lambda }_{n}L)=(-1{)}^{n-1}$ cancels the alternating coefficient sign, leavingEvery summand is positive. The endpoint temperature never drops below $T_0$; it decreases towards $T_0$ as $t\to \infty$ and starts at $T(L,0)=2T_0$.

For completeness, the Sturm-Liouville eigenfunction expansion gives the initial profile in the $L^2$ sense. Here the coefficients are absolutely summable, so the Weierstrass M-test also gives uniform convergence on $[0,L]\times [0,\infty )$, and the initial values are recovered everywhere. For positive times the exponential factors justify all required termwise derivatives, giving a classical solution. There is, however, an initial incompatibility with a Neumann boundary condition at $(L,0)$: the initial slope is $T_0/L$, while the insulated Neumann boundary condition requires zero slope for $t>0$. The temperature is continuous to the initial edge, but $T_x$ cannot extend continuously to that corner. One must not differentiate the initial series term by term at $x=L$.

With $x=2z-1$, the chain rule gives $T_{xx}=V_{zz}/4$. ThusSeparation of variables gives the eigenfunctions $\sin (n\pi z)$ with time factors $e^{-n^2{\pi }^{2}t/4}$. The Fourier sine coefficients of the initial constant areOnly odd modes survive, soReturning to $x$, $\sin ((2k+1)\pi (x+1)/2)=(-1{)}^k\cos ((2k+1)\pi x/2)$, henceFor every $t>0$ the exponential coefficients give uniform convergence of the differentiated series on later-time compact sets, verifying the PDE and boundary conditions. As $t\downarrow 0$, the solution tends to $T_0$ at every interior point and in L2 norm, while remaining zero at the endpoints. The expression is visibly even in $x$, consistent with the corrected symmetry argument for $b=0$.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

Set $u=X(x)Y(y)$ in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in $x$ has the constant mode $X_0=1$ with ${\lambda }_0=0$, and the eigenfunctions $X_n=\cos (nx)$ with ${\lambda }_n=n^2$, $n\ge 1$. Indeed, integration by parts makes $\lambda$ nonnegative; for $\lambda >0$ the first Neumann boundary condition removes the sine term and the second requires $\sqrt{\lambda }$ to be an integer. For $\lambda =0$, both derivative conditions make $X$ constant.

Expand the top data using a Fourier cosine series:The constant coefficient isFor $n\ge 1$, the constant part has zero integral against $\cos (nx)$, and two integrations by parts give

Since the bottom value is zero, a convenient factor depending on $y$ for the nonconstant modes is $\sinh (n(y+1))$, which vanishes at $y=-1$. Its value at $y=1$ is $\sinh (2n)$. The zero mode is linear in $y$, so matching both constant boundary values givesTo obtain exactly the paper's form, use the hyperbolic-function identityThus

Each product of hyperbolic sine or hyperbolic cosine with $\cos (nx)$ has $u_{xx}+u_{yy}=0$. The zero mode is also a harmonic function. On $-1\le y\le 1$, the factor $\sinh (n(y+1))/\sinh (2n)$ lies between zero and one. Since $\sum n^{-2}$ converges, the Weierstrass M-test gives uniform convergence on the closed rectangle. The series therefore gives zero at the bottom and the continuous Fourier cosine series of ${\pi }^2-x^2$ at the top, including its endpoints. On compact subsets with $y<1$, the factors decay exponentially with $n$, justifying termwise partial derivatives. In particular $u_x=0$ on both open vertical edges because every differentiated cosine factor vanishes there. The solution has finite Dirichlet energy: Fourier orthogonality gives the $n$th nonconstant mode's energy as $(\pi /2)n{g}_n^2\coth (2n)=O(n^{-3})$, whose sum is finite. Here $\coth$ is the hyperbolic cotangent. Thus a proves uniqueness in that class.

There is a corner compatibility condition for mixed boundary data in the printed wording. The top data have $g^{\prime }(\pi )=-2\pi$, but the paper writes $u_x(\pi ,y)=0$ even at $y=1$. The series solves the interior equation and all the open-edge boundary conditions; the literal derivative condition at $(\pi ,1)$ is incompatible with the top data. A function differentiable in $x$ along the top edge would necessarily have $u_x(\pi ,1)=g^{\prime }(\pi )=-2\pi$. Hence it cannot simultaneously satisfy the printed zero derivative at that corner, and there is no solution with all those literal corner requirements. The standard interpretation of this mixed boundary condition imposes the Neumann boundary condition on the open sides and the Dirichlet boundary condition continuously on the horizontal edges. Differentiating the series term by term on the top edge would be invalid at this corner; the uniform convergence of the undifferentiated series does not license that operation.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.