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For Picard iteration $y_{n+1}(x)=b+{\int }_a^{x}f(s,y_n(s))ds$, $y_0=b$, assume $|f|\le M$ and a Lipschitz constant $L$ on a rectangle containing every iterate. The first difference is at most $M|x-a|$. Integrating the preceding bound multiplied by $L$ proves the displayed estimate by mathematical induction. Its uniform bounds sum to a convergent exponential series for every finite $Lh$, so the Weierstrass M-test gives uniform convergence without requiring $Lh<1$.

For two arbitrary initial functions $y,z$ in a function set preserved by the Picard iteration operator $T$, the same induction givesThus some iterate is a contraction mapping on any fixed finite interval for which every iterate stays in that set. The iterated contraction argument gives a unique fixed point of $T$ on a nonempty complete metric space, without requiring $Lh<1$. For a rectangle with field bound $M$ and vertical radius $k$, $Mh\le k$ supplies the required self-mapping property.

The period-two extension of $1-|x|$ on $[-1,1]$ is even. Its Fourier cosine series has $a_0=1$ and $a_n=2{\int }_0^1(1-x)\cos (n\pi x)dx=2[1-(-1{)}^n]/(n^2{\pi }^2)$ by integration by parts. This proves the displayed series, with only odd nonconstant harmonics. Continuity and piecewise smoothness give pointwise equality through the Dirichlet-Jordan convergence theorem; the summable coefficient magnitudes also imply uniform convergence by the Weierstrass M-test. At $x=0$, it gives ${\sum }_{m\ge 0}(2m+1{)}^{-2}={\pi }^2/8$.

On $0<x<L$ with zero-temperature endpoints, let the initial temperature rise linearly from zero to height $H$ at $L/2$ and then fall linearly to zero. Integration by parts on the two halves gives Fourier sine series coefficients $8H\sin (n\pi /2)/({\pi }^2{n}^2)$. Under the heat equation $T_t=D{T}_{xx}$ with $D>0$, each coefficient decays by $e^{-D(n\pi /L{)}^{2}t}$, giving the displayed solution. Only odd spatial modes occur; absolute coefficient summability proves uniform convergence for $t\ge 0$ via the Weierstrass M-test. This spatial profile also occurs in a midpoint-plucked fixed-end string, where the time factors oscillate instead of decaying.

Let $S(x)={\sum }_{n\le x}{a}_n$ satisfy $|S(x)|\le C{x}^{\alpha }$ for $x\ge 1$. The Abel summation formula gives $A(s)=s{\int }_1^{\infty }S(t)t^{-s-1}dt$ when $\mathrm{Re}s>\alpha$, since its finite-sum boundary term tends to zero. Split this integral into blocks $I_m(s)={\int }_m^{m+1}S(t)t^{-s-1}dt$. For every $\delta >0$, the estimate $|I_m(s)|\le C{m}^{-1-\delta }$ holds uniformly on $\mathrm{Re}s>\alpha +\delta$. Each block is an entire function, by differentiation under the finite-range integral on compact sets. The Weierstrass M-test makes their sum locally uniformly convergent and therefore holomorphic; multiplication by $s$ preserves holomorphy.

This proof distinguishes uniform convergence of the block sum on a whole half-plane from local uniform convergence after multiplication by $s$. The latter is all that holomorphy needs. In particular, a bound ${\sum }_{n\le x}\mu (n)=O(x^{\alpha })$ gives a holomorphic continuation of the reciprocal Riemann zeta function to this half-plane and excludes zeta zeros there, with the pole at one treated by cancellation.

The first iterate isOn the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\le 102$, and $|F(x,y)-F(x,v)|\le 20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\le h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.

Successive differences satisfy, by induction,The Weierstrass M-test proves uniform convergence on this local interval. The uniform limit is continuous, and the Lipschitz estimate allows passage to the limit inside the integral, giving the displayed integral equation. Differentiation gives $y^{\prime }=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local uniqueness.

Direct differentiation gives $u^{\prime }={\sec }^2(x+\pi /4)=1+u^2$, $u(0)=1$, and $u\to +\infty$ as $x\uparrow \pi /4$. On the common positive existence interval put $w=y-u$. It satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula givesThus $y\ge u$. Let $x_{\ast }>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y^{\prime }\ge 1$. If $x_{\ast }$ were finite and $y$ remained bounded, its finite limit and the smooth right side would allow local continuation, contradicting maximality. Comparison with $u$ prevents existence through $\pi /4$. ConsequentlyThere is no contradiction with Picard's theorem, which gives local existence and uniqueness. Its box estimate gives only a short guaranteed interval, and the Lipschitz constant for $y^2$ is not globally bounded in $y$.

On $[-a,a]$ let $F=\Vert f{\Vert }_{\infty }$ and bound $|K|$ by $M$ on the compact integration region. Write $Th(x)={\int }_0^{x}K(x,s)h(s)ds$. The iteration gives $y_n={\sum }_{j=0}^n{T}^{j}f$. Induction, integrating absolute values also when $x<0$, givesThe exponential series is summable uniformly for $|x|\le a$. The Weierstrass M-test therefore proves uniform convergence to a continuous function $y$. Also $\Vert T(y_n-y){\Vert }_{\infty }\le Ma\Vert y_n-y{\Vert }_{\infty }\to 0$, so passing to the limit yields $y=f+Ty$.

For the indicated kernel, twice differentiating its integral givesConversely integrating the differential equation twice, with these initial conditions, gives the integral equation. This is valid also for negative $x$, either by oriented integration or by differentiating the double-integral identity.

For every finite $a$ the continuous kernel $(x-s)s$ meets the preceding hypotheses. Solutions on overlapping intervals agree: if $d=Td$ is their difference, iteration gives $\Vert d{\Vert }_{\infty }\le \Vert d{\Vert }_{\infty }(Ma{)}^j/j!$ for every $j$, forcing $d=0$. The solutions therefore patch to one solution defined on all of $\mathbb{R}$.

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence

With the classical real coefficientsthe partial sum is $s_n^f=a_0/2+{\sum }_{j=1}^n(a_j\cos jx+b_j\sin jx)$. It is the orthogonal projection onto the first trigonometric modes. Completeness of the trigonometric orthonormal system therefore gives $\Vert s_n^f-f{\Vert }_2\to 0$.

For $f\in C^1$ periodic, integration by parts has zero endpoint terms and gives $a_j(f^{\prime })=j{b}_j(f)$, $b_j(f^{\prime })=-j{a}_j(f)$, and $a_0(f^{\prime })=0$. ThusFor periodic $C^2$ functions, another integration by parts gives $a_j(f)=-a_j(f^{\prime \prime })/j^2$ and $b_j(f)=-b_j(f^{\prime \prime })/j^2$. Riemann-Lebesgue makes the numerator sequences tend to zero, in particular bounded. The Weierstrass M-test and $\sum j^{-2}<\infty$ prove absolute and uniform convergence. The resulting continuous sum agrees with $f$ in $L^2$ by the first result. Two continuous functions equal almost everywhere are equal everywhere, so the pointwise sum is $f$ for every real $x$.

For the polynomial example the periodic extension is $C^2$: the function, its first derivative and its second derivative match at the endpoints. It is even, so all sine coefficients vanish. Direct integration givesFor example, integration by parts gives ${\int }_0^{\pi }{x}^2\cos (jx)dx=2\pi (-1{)}^j/j^2$ and ${\int }_0^{\pi }{x}^4\cos (jx)dx=4{\pi }^3(-1{)}^j/j^2-24\pi (-1{)}^j/j^4$, from which the cancellation of the $j^{-2}$ terms follows. ThereforeAt $x=0$, the left side vanishes, yielding

For $n\le x$, write $F(n)=F(x)-{\int }_n^x{F}^{\prime }(t)dt$, multiply by $f(n)$ and sum. Interchanging a finite sum with the integral gives the Abel summation formulaApply it with $F(t)=t^{-s}$ and $S(t)={\sum }_{n\le t}{a}_n$. If $\sigma >\alpha$, the boundary term is $O(x^{\alpha -\sigma })\to 0$ and ${\int }_1^{\infty }|S(t)t^{-s-1}|dt$ converges. Therefore the Dirichlet series converges andChoose $C$ with $|S(t)|\le C{t}^{\alpha }$ for $t\ge 1$. For $\sigma >\alpha +\delta$,The Weierstrass M-test proves uniform convergence of $\sum I_m$ on this entire open half-plane. For fixed $s$, the same estimate with its actual $\sigma$ gives $I_m(s)=O(m^{\alpha -1-\sigma })$. Each $I_m$ is an entire function, since its integral has bounded positive $t$ range and can be differentiated in $s$ under an integrable bound on each compact set. Thus the locally uniform sum is holomorphic, and so is its product with $s$. This gives holomorphy from polynomial growth of arithmetic partial sums and proves holomorphy of $A$ for $\mathrm{Re}s>\alpha$. Multiplication by the unbounded factor $s$ does not assert uniform convergence of $A$ on a whole unbounded half-plane; local uniform convergence is what the holomorphy argument needs.

With $a_n=\mu (n)$ and the hypothetical bound, $A_{\mu }(s)$ is holomorphic for $\mathrm{Re}s>3/4$. On $\mathrm{Re}s>1$, the Euler product gives $\zeta (s)A_{\mu }(s)=1$. The identity theorem extends this as a meromorphic identity throughout the connected half-plane $\mathrm{Re}s>3/4$. Hence $\zeta$ has no zeros in that half-plane: at any zero, the product with the holomorphic $A_{\mu }$ would be zero. The pole at one is not a zero; $A_{\mu }$ has a compensating zero there.

The extension. Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal topological space can be separated by a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$.

For the bounded Tietze extension theorem, first suppose $g:C\to [-M,M]$ is continuous. The subsets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are closed in $X$, because $C$ is closed. Urysohn's lemma supplies a continuous function $h:X\to [-M/3,M/3]$ with $h=-M/3$ on $A$ and $h=M/3$ on $B$. Directly checking the two outer ranges and the middle range of $g$ givesEmpty level sets cause no difficulty: take the appropriate constant, or zero if both are empty. This is the Urysohn two-thirds approximation.

Apply this successively to the residual on $C$, beginning with $f$ and $M_0=1$. We get $h_n$ with $\Vert h_n{\Vert }_{\infty }\le \frac{1}{3}(\frac{2}{3}{)}^n$ and residual after the first $n$ terms at most $(2/3{)}^n$. The Weierstrass M-test gives uniform convergence of $F={\sum }_{n\ge 0}{h}_n$. By the uniform limit theorem, $F$ is continuous; the residual bound proves $F{|}_C=f$, and the sum of the norm bounds is one. Thus

Closed sets as exact zero sets. If $X$ is a perfectly normal space, write a closed $C$ as ${\bigcap }_{n\ge 0}{U}_n$ with each $U_n$ open. Apply Urysohn's lemma to $C$ and $X\setminus U_n$ to obtain $f_n:X\to [0,1]$ vanishing on $C$ and equal to one on $X\setminus U_n$. The uniformly convergent seriesIndeed all summands vanish on $C$; outside $C$, at least one summand is strictly positive. The uniform limit theorem proves continuity. This realizes the closed G-delta set as a zero set construction.

Conversely, suppose each closed set is the zero set of a continuous function. For disjoint closed $C,D$, select $f,g:X\to [0,1]$ with zero sets $C,D$. Their sum never vanishes, andThe quotient is continuous. This also proves normality by inverse images of disjoint endpoint neighborhoods. Each zero set is a G-delta set, since $f^{-1}(0)={\bigcap }_{n\ge 1}{f}^{-1}([0,1/n))$; thus this property characterizes perfectly normal spaces among T1 spaces.

Subspaces. Let $Y\subseteq X$ and let $A,B$ be disjoint closed subsets of $Y$. Put $C={\overline{A}}^X$ and $D={\overline{B}}^X$. Since $C\cap Y=A$ and $D\cap Y=B$, the intersection $C\cap D$ misses $Y$. Choose continuous functions $f,g$ on $X$ whose zero sets are $C,D$. On the open neighborhood $W=X\setminus (C\cap D)$ of $Y$, the same quotient $f/(f+g)$ is continuous and is zero on $A$, one on $B$. Its inverse images of $[0,1/3)$ and $(2/3,1]$, restricted to $Y$, are disjoint open neighborhoods. Therefore every subspace is a normal topological space: perfect normality implies hereditary normality.

A strict counterexample to the converse. Let $D$ be an uncountable discrete space, and let $X=D\cup \{\infty \}$ be its one-point compactification. Its neighborhoods of $\infty$ have finite complement in $D$, because compact subsets of a discrete space are finite. The space is compact Hausdorff, hence a T1 space and normal. A subspace missing $\infty$ is discrete. A subspace containing $\infty$ is closed in $X$, since its complement is a subset of the open discrete part; it is therefore compact Hausdorff and normal. Thus all subspaces are normal.

But $\{\infty \}$ is not a G-delta set. A countable intersection of its open neighborhoods omits at most countably many elements of $D$, and cannot omit all of uncountable $D$. Hence $X$ is not a perfectly normal space. This is the uncountable discrete one-point compactification counterexample.

For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

For the direct contraction mapping theorem proof with the usual supremum norm, take the complete metric spaceand the Picard iteration map $Tz(x)=b+{\int }_a^{x}f(s,z(s))ds$. The estimates areThus this proof requiresThe first inequality makes $T$ preserve $\mathcal{B}$; the strict second inequality makes it a contraction mapping. Its unique fixed point is the desired solution of a differential equation.

This is more restrictive than the preceding Weierstrass M-test argument: a single application of $T$ must shrink every difference in the supremum norm, whereas repeated integration in the other proof produces the factorial $(n+1)!$ and convergence even when $Lh\ge 1$. If $L=0$, the additional condition is automatic. These are conditions for the two stated direct proofs; they are not necessary bounds on the actual existence interval.

The bound $0\le d_n\le 2^{-n-1}$ proves uniform convergence of the series by the Weierstrass M-test. Each summand is continuous and satisfies $d_n(x+1)=d_n(x)$, so its sum $t$, the Takagi function, is continuous and has period one.

For $n\ge m$, both $2^n{u}_m$ and $2^n{v}_m$ are integers, giving $d_n(u_m)=d_n(v_m)=0$. ThereforeThe corners of $d_n$ occur on the grid $2^{-n-1}\mathbb{Z}$. When $n<m$, that grid is contained in $2^{-m}\mathbb{Z}$, so the interior of $[u_m,v_m]$ contains no corner of $d_n$. Its slope on this interval is consequently $+1$ or $-1$.

The interval $[u_{m+1},v_{m+1}]$ is a half of $[u_m,v_m]$. All the old slopes persist on this half, and the new summand contributes one more slope in $\{\pm 1\}$. ThusIf $t$ had a finite derivative $L$ at $x_0$, write $t(y)=t(x_0)+L(y-x_0)+r(y)$ with $r(y)=o(|y-x_0|)$. Since $u_m\le x_0<v_m$ and both endpoints tend to $x_0$,Indeed $|v_m-x_0|+|u_m-x_0|=v_m-u_m$; when $u_m=x_0$, its remainder is zero. This is the secants straddling a differentiability point argument, including dyadic $x_0$. But the displayed increments prevent $S_m$ from converging. Hence $\boxed{t\text{ is nowhere differentiable}}$, using periodicity to cover all real $x_0$.

For the distributional derivative, set $r_n(x)=s(2^{n}x)$, using $s$ from part (iii). The change of variable in the distributional derivative pairing gives $d_n^{\prime }=r_n$; these are the Rademacher functions. Uniform convergence implies weak convergence of distributions, and the continuity of distributional differentiation allows termwise differentiation in that topology. Explicitly,This series assertion is a distributional convergence statement. Its remainder satisfies the concrete estimateIt supplies a well-defined distributional derivative even though the ordinary derivative exists nowhere.

The $2\pi$ periodization of a Schwartz function isFor every nonnegative integer $j$, the Schwartz seminorms give $|{\varphi }^{(j)}(x+2\pi k)|\le C_j(1+|k|{)}^{-2}$ uniformly for $x\in [0,2\pi ]$. The Weierstrass M-test gives absolute and uniform convergence there of the defining series and all its differentiated series. Termwise differentiation proves that $P\varphi$ is a smooth function, with $(P\varphi {)}^{(j)}=P({\varphi }^{(j)})$. Reindexing the absolutely convergent series proves $P\varphi (x+2\pi )=P\varphi (x)$, so it is a periodic function.

For an explicit smooth periodic partition of unity, choose the bump functionIt is a test function, positive on $[-\pi ,\pi ]$. The smooth function $S(x)=P\eta (x)$ is $2\pi$ periodic and strictly positive, since every $x$ has a translate in $[-\pi ,\pi ]$. SetDivision preserves smoothness because $S>0$, and the support remains contained in $[-2\pi ,2\pi ]$.

Put $h(x)=({\alpha }^2+x^2{)}^{-1}$ and form its period-one periodization of an integrable functionThe function $h$ is not a Schwartz function, so justify the needed Poisson summation formula directly. For every $j\ge 0$, its derivative satisfies $h^{(j)}(x)=O(|x{|}^{-2-j})$ as $|x|\to \infty$. Thus the Weierstrass M-test gives uniform convergence on $[0,1]$ of this series and each differentiated series. It follows that $H$ is a smooth period-one periodic function.

Absolute integrability and Fubini's theorem allow its Fourier coefficients to be computed by splitting the real line into unit intervals:These Fourier coefficients are absolutely summable. Their Fourier series defines a continuous periodic function with the same Fourier coefficients as $H$; uniqueness of Fourier coefficients in L1 identifies the two almost everywhere, and continuity then identifies them everywhere. Evaluating at zero gives the rigorously justified Poisson summation formulaRearranging provesThe last series is a geometric series with sum $(e^{2\pi \alpha }-1{)}^{-1}$. Equivalently, the left side is $\pi \coth (\pi \alpha )/(2\alpha )-1/(2{\alpha }^2)$, where $\coth$ is the hyperbolic cotangent.

The exponential series gives $x^x=e^{x\log x}={\sum }_{k=0}^{\infty }(x\log x{)}^k/k!$. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test proves uniform absolute convergence, using the summable bounds $e^{-k}/k!$. Therefore one can integrate the series term by term over this finite interval. Using part i,This is the alternating identity in sophomore's dream. The same estimate also proves absolute convergence of the integrated series; no cancellation is needed to justify the interchange.

Fix $z_0\in E$ and $ϵ>0$. Choose $N$ so that ${\sup }_E|f_N-f|<ϵ/3$, and then choose $\delta >0$ so that $|f_N(z)-f_N(z_0)|<ϵ/3$ for $z\in E$ with $|z-z_0|<\delta$. The triangle inequality gives $|f(z)-f(z_0)|<ϵ$. This proves the uniform limit theorem on the relative topology of $E$, whether or not $E$ is open.

The Weierstrass M-test states that if $|u_n(z)|\le M_n$ for every $z\in E$ and $\sum M_n<\infty$, then $\sum u_n$ converges uniformly and absolutely on $E$. Indeed, the supremum of a tail is at most the corresponding tail of $\sum M_n$, proving the uniform Cauchy criterion.

For the given secant function terms, a denominator can vanish only at $z=n(2k+1)$, an integer. Thus every individual term is continuous off the integers. Fix $z_0\notin \mathbb{Z}$, and choose a closed disk $K$ about $z_0$ containing no integer. Let $R={\sup }_K|\mathrm{Re}z|$ and take $N>2R$. For $n\ge N$, $|\pi \mathrm{Re}z/(2n)|<\pi /4$; the permitted cosine inequality givesThe Weierstrass M-test makes the tail uniformly convergent on $K$. The finitely many earlier terms are continuous there, so the uniform limit theorem makes their sum continuous at $z_0$. Therefore $f$ is continuous on $\mathbb{C}\setminus \mathbb{Z}$. The argument is local; a global M-test on the whole punctured plane would fail near its poles.

Let $K\subset \mathbb{C}\setminus \mathbb{Z}$ be compact and contained in $|z|\le R$. For $|n|>2R$, $|z-n|\ge |n|/2$, hence $|(z-n{)}^{-2}|\le 4/n^2$ uniformly on $K$. The tail is therefore absolutely and uniformly convergent by the Weierstrass M-test. Its finitely many remaining summands are holomorphic on a neighborhood of $K$.

The theorem that a locally uniform limit of holomorphic functions is holomorphic now proves that the sum is analytic on $\mathbb{C}\setminus \mathbb{Z}$. Absolute convergence also permits a shift of the integer index:Thus the function has period one.

The coefficient sum is finite, so the Weierstrass M-test makes the cosine series uniformly convergent and $f$ continuous. Put $Q=5^{m+1}$ and $A_m={\sum }_{k=m+1}^{\infty }{a}_k$. The indicated partial sum belongs to ${\mathcal{T}}_n$ because its largest frequency is $5^m\le n$. Its tail satisfiesand equality holds at $x=0$. Thus its supremum norm error is exactly $A_m$.

At the $2Q$ points $x_j=j\pi /Q$, $0\le j<2Q$, every omitted frequency is an odd integer multiple of $Q$. Indeed for $k\ge m+1$,Uniform convergence permits summing these values, givingThese are cyclically alternating extrema. Since $n<Q$, there are $2Q\ge 2n+2$ of them, so the trigonometric Chebyshev alternation theorem proves that the partial sum is the unique best approximant. Consequently the positive lacunary trigonometric series hasThe positivity of every tail coefficient makes the same tail attain its full coefficient sum with alternating signs; a triangle bound alone would not establish optimality.

Fix $x_0\in E$ and $\epsilon >0$. By uniform convergence, choose $N$ so that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in E$. By continuity of $f_N$ relative to $E$, there is $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality givesThis proves the uniform limit theorem without requiring $E$ to be open or closed.

For the series, fix an arbitrary $x_0\in (0,1]$ and put $a=x_0/2>0$. On $[a,1]$ each summand $n^{-1-x}$ is continuous andThe Weierstrass M-test gives uniform convergence of the series on $[a,1]$. Its partial sums are continuous, so the uniform limit theorem makes its sum continuous on that interval, in particular at $x_0$ with the appropriate relative topology if $x_0=1$. Since $x_0$ was arbitrary, the sum is continuous on $(0,1]$.

This is local uniform convergence, not uniform convergence on the entire half-open interval. Indeed, for any finite partial sum length $N$, its tail as $x\downarrow 0$ is bounded below by arbitrarily long tails of the divergent harmonic series. There is no continuity claim at the excluded endpoint $0$.