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The sequence has uniform convergence on $X$ if there is a function $f:X\to \mathbb{R}$ such thatThe same $N$ must work for every $x$. Equivalently, the supremum of the error over $X$ tends to zero. In pointwise convergence, $N$ is allowed to depend on $x$.

For each fixed $0<x<1$, $x^n\to 0$, so the pointwise limit is $f(x)=1/(1-x)$. Howeverfor every fixed $n$. Hence the error has infinite supremum on $(0,1)$ and the convergence is not uniform. This does not contradict uniform convergence on any smaller closed interval $[0,r]$, $r<1$, where the error is at most $r^n/(1-r)$.

The function is even, so its Fourier series has $b_n=0$. In the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$,Twice applying integration by parts gives, for $n\ge 1$,Therefore the periodic Fourier representation iswith periodic continuation. The periodic extension is continuous, including the identified endpoints, and piecewise smooth, so the Fourier series converges to it. The coefficients also give absolute convergence and uniform convergence by comparison with $\sum n^{-2}$.

At $x=\pi$, the cosine cancels the alternating coefficient sign; at $x=0$, it is identically one. These evaluations yieldThe first identity solves the Basel problem.

Take ${\Omega }^{\prime }$ nonempty and compactly contained in $\Omega$. The argument is a compactness proof of a nonnegative elliptic solution estimate, with the global Schauder estimate retaining control all the way to the boundary. The operator and the Hölder space exponent are fixed throughout.

Suppose the estimate fails. For each integer $k$ there is an admissible nonnegative solution $u_k$, with forcing $f_k$, such thatPut $v_k=u_k/M_k$ and $g_k=f_k/M_k$. Then $0\le v_k\le 1$, ${\max }_{\overline{\Omega }}{v}_k=1$, $v_k=0$ on the boundary, $L{v}_k=g_k$, andGlobal elliptic regularity and the global Schauder estimate for zero Dirichlet boundary condition giveThis estimate includes the $C^0$ term, so it does not require invertibility of $L$ or a sign restriction on $c$. The compact embedding of Hölder spaces, or the Arzela-Ascoli theorem applied through second partial derivatives, gives a subsequence converging in $C^2(\overline{\Omega })$ to $v$. Thus $v\ge 0$, $Lv=0$, $v=0$ on the boundary, and $\max v=1$.

Choose $x_k\in {\Omega }^{\prime }$ with $v_k(x_k)\le {\inf }_{{\Omega }^{\prime }}{v}_k+1/k$. A further subsequence has $x_k\to x_{\ast }\in \overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence yields $v(x_{\ast })=0$. Part (b) says that a nonnegative homogeneous solution on the connected domain is either zero everywhere or strictly positive inside. Both alternatives contradict, respectively, $\max v=1$ and $v(x_{\ast })=0$. ThereforeThe interior infimum and the forcing control the global supremum. Global boundary control is crucial here: an interior-only compactness argument could lose the normalized maximum at the boundary.

Set $\xi =hk$. For the discrete Fourier mode, the displacement difference in the $l$th neighbor pair carries $e^{il\xi }-1$, andThus the infinite system reduces toThe gravitating lattice Fourier kernel $F$ is nonnegative, even and $2\pi$-periodic, with $F(0)=0$. The uniform convergence and absolute convergence of its differentiated series givesTo prove the maximum, it is not sufficient to maximize each cosine term separately: the even terms do not reach their individual maxima at $\pi$. Instead use the convergent integral representation $l^{-2}={\int }_0^{\infty }t{e}^{-lt}dt$. Summing the geometric series for sines givesFor $0<\xi <\pi$, the denominator is positive and $\sin \xi >0$, so $F^{\prime }>0$. For $\pi <\xi <2\pi$, $F^{\prime }<0$. The interchange is justified by ${\sum }_l{\int }_0^{\infty }t{e}^{-lt}|\sin (l\xi )|dt\le {\sum }_l{l}^{-2}<\infty$. HenceTheir value iswhere $\zeta$ is the Riemann zeta function. This maximizing Fourier mode alternates the signs of neighboring particle displacements.

For the constant monomial $f_0=1$, the binomial theorem gives $B_n(f_0)(x)=(x+1-x{)}^n=1$. For $f_1$, the binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ givesThus $\boxed{B_n(f_0)=1,B_n(f_1)=x}$, including $x=0,1$ by evaluation of the polynomials.

For $m\ge 1$ and $n\ge 2$, substitute the preceding coefficient identity into the Bernstein polynomial sum and write $j=k-1$. This yields the Bernstein monomial recurrenceThe coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n{)}^{m-1}\to 1$ and the others tend to zero. The weights in any Bernstein polynomial sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le 1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x{x}^{m-1}=x^m$: the $\ell =0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the supremum norm proves uniform convergence on $[0,1]$.

As an independent quantitative bound, take a binomial distribution variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb{E}[(K/n{)}^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the Cauchy-Schwarz inequality and the variance of $K/n$ giveFor $m=0$ the error is exactly zero. Finally the requested sum equals $2^m{B}_{2n}(f_m)(1/2)$, so

True. Define $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one. The restriction of $f$ to $[-n,n]$ is uniformly continuous by the Heine-Cantor theorem; its modulus of continuity combined with $|c_n(x)-c_n(y)|\le |x-y|$ proves uniform continuity of $f_n$ on all of $\mathbb{R}$. For each fixed $x$, once $n\ge |x|$ we have $f_n(x)=f(x)$. HenceThis pointwise approximation by uniformly continuous functions even gives eventual equality on every fixed bounded interval. It need not give uniform convergence on the whole line, as the bounded counterexample in part (i) already demonstrates.

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.

No: local uniform convergence need not be global uniform convergence. For a bounded counterexample use the continuous functions $f_n(x)=\min (1,|x|/n)$ and $f(x)=0$. On any fixed interval $[-M,M]$, for $n>M$,Thus there is locally uniform convergence to zero. On all of $\mathbb{R}$, however, ${\sup }_x|f_n(x)-f(x)|=1$ for every $n$, so uniform convergence fails. The cutoff needed to control the tails cannot be chosen independently of the interval.

First prove the hinted Caccioppoli inequality. Testing the weak formulation with $u_k{\phi }^2$ is legitimate by approximation in the zero-boundary Sobolev space. Let ${\Lambda }_{\ast }=\mathrm{esssup}\Vert a(x){\Vert }_{\mathrm{op}}\le n\Lambda$. Thenby Young inequality. HenceNo derivative of the measurable coefficient matrix is taken.

Set $w_k=u_k/{\sup }_{B_1}|u_k|$. The denominator is finite and positive because $u_k$ is continuous on the compact closure and nonzero. Each $w_k$ solves the same linear equation and $|w_k|\le 1$. For every $\theta <1$, choose a cutoff function equal to one on $B_{\theta }$ and supported in a slightly larger interior Euclidean ball. The energy bound gives a uniform $W^{1,2}(B_{\theta })$ bound.

Take radii ${\theta }_j\uparrow 1$. Weak compactness and a diagonal subsequence argument produce one subsequence converging weakly in $W^{1,2}(B_{{\theta }_j})$ for every $j$, with consistent restrictions defining $v\in W_{\mathrm{loc}}^{1,2}(B_1)$. Part (b) and the Arzela-Ascoli theorem allow a further subsequence converging uniformly on $\overline{B_{1/4}}$. Its uniform limit agrees almost everywhere with the weak limit and supplies its continuous representative there.

For every smooth compactly supported test function, choose $j$ containing its support. Since $a^{ij}{D}_i\phi \in L^2$ and $D{w}_k$ converges weakly there,Thus the compactness of normalized weak elliptic solutions givesThe diagonal step is needed to retain the equation beyond the small Euclidean ball of uniform convergence. The limit need not be nonzero: for the Laplacian in dimension at least two, the normalized harmonic functions $\mathrm{Re}(x_1+i{x}_2{)}^k$ tend uniformly to zero on each smaller Euclidean ball.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.

Use the Fourier transform normalization $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. A precise version of the Poisson summation formula is that, for every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. Here being a Schwartz function means being smooth with ${\sup }_x|x^a{f}^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.

Form the periodization of a Schwartz function $P_f(x)={\sum }_{n}f(x+n)$. Every differentiated series has uniform convergence on $[0,1]$, by rapid decay, so $P_f$ is a smooth periodic function of period one. Its $m$th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of $|m|$. Thus the Fourier series converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.

For $t>0$, scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation

For $\mathrm{Re}s>1$, termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive $n$ contributes $\Gamma (s/2)(\pi n^2{)}^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.

Split the integral at one. On $(0,1)$, the Jacobi theta function transformation writes $\theta (t)-1=t^{-1/2}-1+t^{-1/2}(\theta (1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of $s$: $\theta (t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a meromorphic continuation of $\Lambda$ to the whole plane, with simple poles at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto 1-s$.

The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at $s=1$, of residue one. The apparent pole at $s=0$ cancels; in fact $\Gamma (s/2{)}^{-1}\sim s/2$ gives $\zeta (0)=-1/2$. This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is an entire function with $\xi (s)=\xi (1-s)$.

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

For each $k$, choose $n_k$ with $\lambda (A\setminus E_{n_k}^{(k)})\le \epsilon 2^{-k-1}$ and set $B={\bigcap }_{k\ge 1}{E}_{n_k}^{(k)}$. Countable subadditivity gives $\lambda (A\setminus B)\le \epsilon /2$. For any tolerance $\delta >0$, choose $k$ with $1/k<\delta$. For all $m\ge n_k$ and every $x\in B$, $|f_m(x)-f(x)|\le 1/k$, proving uniform convergence on $B$.

If the functions are Borel measurable, $B$ itself is Borel and we can take $A_{\epsilon }=B$. If “measurable” is interpreted in the completed Lebesgue sense, use regularity of Lebesgue measure to choose a Borel subset $A_{\epsilon }\subseteq B$ differing from $B$ by a null set; the same measure bound and uniform convergence hold. Thus the requested Borel conclusion is valid under either convention. This proves the finite-measure Egorov theorem.

For an infinite-measure counterexample take $A=[0,\infty )$ and $f_n=1_{[n,\infty )}$. It converges pointwise to zero. Any subset of $A$ with complement of finite measure intersects every infinite tail $[n,\infty )$, so the supremum of $f_n$ there is one for every $n$. Hence finite measure is essential to the conclusion.

True. By the Weierstrass approximation theorem, choose a polynomial $Q_n$ with ${\sup }_{|t|\le n}|Q_n(t)-f(t)|<1/n$. For each fixed $t$, this bound applies once $n\ge |t|$, so $Q_n(t)\to f(t)$. This proves pointwise convergence; the same construction even gives uniform convergence on every closed interval.

False. If uniform convergence held, some polynomial $P_n$ would satisfy ${\sup }_{t\in \mathbb{R}}|P_n(t)-\sin t|<1$. Such a polynomial would be bounded on $\mathbb{R}$, hence constant: a nonconstant polynomial grows without bound in absolute value along a real half-line. For any constant $c$, however,a contradiction.

True. Given $\epsilon >0$, uniform continuity supplies $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. For all sufficiently large $n$, $1/n<\delta$, and thenThis is precisely uniform convergence of $f_n$ to $f$.

Let $u_j\in C_c^{\infty }({\mathbb{R}}^n)$ tend to $u$ in the Sobolev space $W^{1,p}$, using density of smooth functions in a Sobolev space. Write $\alpha =1-n/p>0$. Applying the Morrey inequality on a cube to a unit cube centered at $x$, and bounding its average by the Holder inequality, givesfor every smooth $v$. Hence $u_j$ is a Cauchy sequence in the supremum norm and has a bounded continuous limit $u^{\ast }$ by uniform convergence. On every finite-measure cube, this limit is also the $L^p$ limit, so $u=u^{\ast }$ almost everywhere.

For $x\ne y$, choose $r=4|x-y|$, so that $y\in Q_r(x)$. The second estimate from part (a), followed by passage to the limit, givesThus $u^{\ast }$ is a Hölder continuous function andIt is the unique continuous representative: two continuous functions agreeing almost everywhere agree everywhere.

For $1\le p<n$, take a smooth cutoff function $\chi$ equal to one near zero and setIts weak derivative has size $O(|x{|}^{-a-1})$ near zero, whose $p$th power is integrable because $p(a+1)<n$. Thus $u\in W^{1,p}$, but it is essentially unbounded in every neighborhood of zero and has no continuous representative. The value assigned at zero is irrelevant.

Choose $x$ at which $Du(x)$ is finite and the $p$-mean version of the Lebesgue differentiation theorem holds:These conditions hold almost everywhere. On a cube about $x$, subtract the affine function and put $v(y)=u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. Its weak derivative is $Dv=Du-Du(x)$ and $v(x)=0$. Approximate $u$ as in part (b) and subtract the same affine function from the approximants. Uniform convergence and convergence of their weak derivatives extend part (a)'s estimate to $v$, givingFor $h\ne 0$, choose $r=4|h|$ and $y=x+h$. It follows thatThis is precisely classical Frechet derivative differentiability, not merely existence of coordinate derivatives. Therefore the differentiability almost everywhere of supercritical Sobolev functions gives