Convergence of Fourier series

Theorem

Convergence of Fourier series

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Statement

Let \( f : \mathbb{R} \to \mathbb{R} \) (or \( \mathbb{C} \)) be \( 2\pi \)-periodic and piecewise \( C^1 \) on \( [-\pi,\pi] \): there is a finite set of points in \( [-\pi,\pi] \) outside which \( f \) is continuously differentiable, and at every point \( x \in \mathbb{R} \) the one-sided limits \( f(x^+) = \lim_{t \to 0^+} f(x+t) \), \( f(x^-) = \lim_{t \to 0^-} f(x+t) \), and the one-sided derivatives \( \lim_{t \to 0^+} \frac{f(x+t)-f(x^+)}{t} \), \( \lim_{t \to 0^-} \frac{f(x+t)-f(x^-)}{t} \) all exist. Define the Fourier coefficients \( a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)\cos(nt)\,dt \) for \( n \geq 0 \), \( b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)\sin(nt)\,dt \) for \( n \geq 1 \), and the partial sums \( S_N f(x) = \frac{a_0}{2} + \sum_{n=1}^{N} \left( a_n \cos(nx) + b_n \sin(nx) \right) \). Then for every \( x \in \mathbb{R} \), \[ \lim_{N \to \infty} S_N f(x) = \frac{f(x^+) + f(x^-)}{2}. \] In particular, if \( f \) is continuous at \( x \), the Fourier series converges to \( f(x) \) there.

Why it matters

Fourier's claim that "any" periodic function is a sum of sines and cosines was, for a century after 1807, a claim without a precise sense of convergence attached to it — and the history of nineteenth-century analysis is largely the history of pinning that sense down. The Dirichlet–Jordan pointwise theorem is the first fully rigorous answer: it isolates exactly the local regularity (a one-sided derivative on each side of \( x \)) that forces the partial sums to home in on the midpoint of the jump, and it does so by a mechanism – localisation plus the Riemann–Lebesgue lemma – that becomes the template for every finer convergence theorem that follows it.

It also matters practically: it is the theorem that licenses solving the heat and wave equations on an interval by separation of variables and then justifying, term by term, that the resulting trigonometric series actually reproduces the initial data at each point – the reason this result sits inside a PDE unit rather than only a pure analysis course.

Hypotheses

Periodicity, \( f(x+2\pi) = f(x) \) for all \( x \). Without periodicity the Fourier coefficients still exist if \( f \) is integrable on \( [-\pi,\pi] \), but \( S_N f \) is itself always \( 2\pi \)-periodic, so it can only converge to a \( 2\pi \)-periodic function; for non-periodic \( f \) the series simply reconstructs the periodic extension, not \( f \) itself outside \( [-\pi,\pi] \).

Piecewise continuity of \( f \) (so the defining integrals for \( a_n, b_n \) exist). Drop this and the coefficients themselves may fail to exist, e.g. \( f(x) = 1/x \) near \( 0 \) inside a period has a non-integrable singularity, so \( a_n, b_n \) are not even defined.

Existence of the one-sided derivatives \( \lim_{t\to0^+} \frac{f(x+t)-f(x^+)}{t} \) and \( \lim_{t\to0^-} \frac{f(x+t)-f(x^-)}{t} \) at the point \( x \) in question (a Dini/Lipschitz-type condition). This is the load-bearing hypothesis. Du Bois-Reymond (1873) constructed a function continuous on the circle whose Fourier series diverges at one specific point precisely because no such one-sided derivative exists there – continuity of \( f\) alone is not enough, only continuity plus this local smoothness controls the tail of the series at that point.

Only finitely many points of discontinuity of \( f \) and \( f' \) in each period (piecewise, not merely pointwise, regularity). If \( f \) is allowed infinitely many jumps accumulating at a point (e.g. \( f \) built from countably many scaled copies of a sawtooth with jump heights \( 1/k^2 \) placed at \( 1/k \)), the argument below still applies pointwise at every point where the one-sided-derivative condition holds, but bounding the coefficients and controlling uniform behaviour near the accumulation point requires extra care that the classical statement does not supply.

Proof

D_N(t) := \sum_{n=-N}^{N} e^{int} = \frac{\sin\!\left((N+\tfrac12)t\right)}{\sin(t/2)}, \qquad t \neq 2k\pi

Definition of the Dirichlet kernel as a partial sum of the geometric series \( \sum_{n=-N}^{N} z^n \) with \( z = e^{it} \); summing the finite geometric series \( \sum_{n=-N}^{N} z^n = z^{-N}\frac{z^{2N+1}-1}{z-1} \) and simplifying with \( z = e^{it} \) using \( e^{i\theta}-e^{-i\theta} = 2i\sin\theta \) gives the closed form. A

S_N f(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x+t)\, D_N(t)\, dt

Substitute the integral formulas for \( a_n, b_n \) into \( S_N f(x) \), write \( \cos(n(x-s)) \) via \( e^{in(x-s)} \), interchange the finite sum and the (fixed-limits) integral, and substitute \( t = s-x \) using periodicity of \( f \) to shift the domain of integration back to \( [-\pi,\pi] \). Each step is a finite linear rearrangement, legitimate since the sum over \( n \) is finite. A

\frac{1}{2\pi}\int_{-\pi}^{\pi} D_N(t)\, dt = 1

Integrate the defining sum \( D_N(t) = \sum_{n=-N}^N e^{int} \) term by term over a full period: every term with \( n \neq 0 \) integrates to \( 0 \) since \( \int_{-\pi}^\pi e^{int}\,dt = 0 \) for integer \( n \neq 0 \), leaving only the \( n=0 \) term, which integrates to \( 2\pi \). A

D_N(-t) = D_N(t), \quad\text{so}\quad \frac{1}{2\pi}\int_{-\pi}^{0} D_N(t)\,dt = \frac{1}{2\pi}\int_{0}^{\pi} D_N(t)\,dt = \frac12

\( D_N \) is a ratio of an odd function (\(\sin\) of an odd multiple) over an odd function \( \sin(t/2)\), hence even; splitting the integral in Step 3 at \( t=0 \) and using evenness gives equal halves, each equal to \( \tfrac12 \) by Step 3. A

S_N f(x) - \frac{f(x^+)+f(x^-)}{2} = \frac{1}{2\pi}\int_0^\pi \big[f(x+t)-f(x^+)\big] D_N(t)\,dt + \frac{1}{2\pi}\int_{-\pi}^0 \big[f(x+t)-f(x^-)\big] D_N(t)\,dt

Multiply the two identities of Step 4 by the constants \( f(x^+) \) and \( f(x^-) \) respectively, add them to get \( \frac{f(x^+)+f(x^-)}{2} = \frac{1}{2\pi}\int_0^\pi f(x^+) D_N(t)\,dt + \frac{1}{2\pi}\int_{-\pi}^0 f(x^-)D_N(t)\,dt \), and subtract this from the identity of Step 2 (itself split at \( t=0 \)). Purely algebraic rearrangement. A

g_+(t) := \frac{f(x+t)-f(x^+)}{\sin(t/2)} \ \ (0 \lt t \leq \pi), \qquad g_+(0^+) := \lim_{t\to0^+} g_+(t) \ \text{exists}

By hypothesis the one-sided derivative \( \ell_+ := \lim_{t\to0^+} \frac{f(x+t)-f(x^+)}{t} \) exists; since \( \frac{t}{\sin(t/2)} \to 2 \) as \( t \to 0^+ \) (standard limit \( \lim_{u\to0} \frac{u}{\sin u} = 1\) with \( u=t/2\)), the product \( g_+(t) = \frac{f(x+t)-f(x^+)}{t}\cdot\frac{t}{\sin(t/2)} \to 2\ell_+ \) as \( t\to0^+ \), so \( g_+ \) extends to a bounded, piecewise-continuous function on \( (0,\pi] \) with a finite right limit at \( 0 \). C

\frac{1}{2\pi}\int_0^\pi \big[f(x+t)-f(x^+)\big]D_N(t)\,dt = \frac{1}{2\pi}\int_0^\pi g_+(t)\,\sin\!\left(\Big(N+\tfrac12\Big)t\right) dt \ \longrightarrow\ 0 \ \text{as } N\to\infty

Substitute \( D_N(t) = \sin((N+\tfrac12)t)/\sin(t/2) \) and the definition of \( g_+ \); since \( g_+ \in L^1(0,\pi) \) by Step 6 (bounded on a finite interval, piecewise continuous), the Riemann–Lebesgue Lemma – for any \( g \in L^1(a,b) \), \( \int_a^b g(t)\sin(\lambda t)\,dt \to 0 \) as \( \lambda \to \infty \) – applies with \( \lambda = N+\tfrac12 \). C

\frac{1}{2\pi}\int_{-\pi}^0 \big[f(x+t)-f(x^-)\big]D_N(t)\,dt \ \longrightarrow\ 0 \ \text{as } N\to\infty

Identical argument to Steps 6–7 applied to \( g_-(t) := \frac{f(x+t)-f(x^-)}{\sin(t/2)} \) on \( [-\pi,0) \), using the left one-sided derivative hypothesis at \( x \) and the Riemann–Lebesgue Lemma again. B

S_N f(x) - \frac{f(x^+)+f(x^-)}{2} \longrightarrow 0 \quad\text{as } N \to \infty

Substitute the two limits of Steps 7 and 8 into the identity of Step 5: both terms on the right vanish, giving convergence of the left side to \( 0 \), which is the claim. B

Result

\lim_{N\to\infty} S_N f(x) = \frac{f(x^+)+f(x^-)}{2}

Reading. At every point, the symmetric partial sums of the Fourier series converge to the average of the left- and right-hand limits of \( f \) – the "midpoint of the jump." At a point of continuity this average is simply \( f(x) \), so the series reproduces the function exactly there.

Scope. Pointwise, one \( x \) at a time: the theorem says nothing by itself about uniform convergence, about convergence in \( L^2 \) or any other norm, or about what happens if the one-sided-derivative hypothesis fails at \( x \) (it may still converge, or may diverge – the theorem is silent). It requires piecewise \( C^1 \) regularity, which is far stronger than what is needed for \( L^2 \) convergence (Carleson's theorem needs only \( f \in L^2 \)) but weaker than what is needed for absolute/uniform convergence (which typically needs continuity plus a Dini or Hölder condition globally, or \( f \in C^1 \) globally without jumps).

Corollaries & converses

Continuity corollary. If \( f \) is piecewise \( C^1 \) and continuous at \( x \), then \( S_N f(x) \to f(x) \); if \( f \) is continuous and piecewise \( C^1 \) on all of \( \mathbb{R} \), the series converges to \( f \) at every point (though not necessarily uniformly – see Fails without).
Uniform convergence corollary. If \( f \) is continuous on \( \mathbb{R} \), \( 2\pi \)-periodic, and piecewise \( C^1 \) with \( f' \in L^2 \), then \( S_N f \to f \) uniformly (proved separately via the Weierstrass M-test applied to \( \sum |a_n|+|b_n| \), bounded using Bessel's inequality for \( f' \) and integration by parts \( a_n(f) = -b_n(f')/n \)); this is a strictly stronger conclusion than the pointwise theorem gives on its own.
Termwise integration. Because convergence holds pointwise a.e. and the partial sums are uniformly bounded in \( L^1 \) (Bessel's inequality), the Fourier series of any piecewise-\(C^1\) periodic \( f \) may be integrated term by term over any interval, recovering the original antiderivative up to a constant.
Converse fails. The converse – "if the Fourier series of \( f \) converges at \( x \) then \( f \) has one-sided derivatives at \( x \)" – is false: Fejér's theorem shows the Cesàro means of the Fourier series of any continuous periodic \( f \) converge uniformly to \( f \), and ordinary convergence can also occur at points where no one-sided derivative exists; the hypothesis is sufficient, not necessary, for pointwise convergence.
Uniqueness (a near-converse worth noting). If two piecewise-\( C^1 \) periodic functions have identical Fourier coefficients, this theorem forces them to agree at every point of joint continuity, hence to agree except possibly at finitely many jump points per period – the Fourier coefficients determine \( f \) essentially uniquely.

Fails without

Drop the one-sided-derivative hypothesis at \( x \): du Bois-Reymond's 1873 example exhibits a function continuous on the circle whose Fourier series diverges (partial sums unbounded) at one particular point, where continuity holds but no one-sided derivative exists; the theorem's proof breaks exactly at Step 6, since \( g_+ \) is no longer guaranteed to be integrable near \( t=0 \), and Riemann–Lebesgue cannot be invoked.
Drop piecewise continuity entirely (allow \( f \in L^1 \) only): Kolmogorov (1926) constructed an \( f \in L^1(\mathbb{T}) \) whose Fourier series diverges at every point; no averaging to a one-sided limit occurs anywhere, showing integrability alone is far too weak a hypothesis for this pointwise conclusion.
Keep the hypotheses but expect uniform convergence for free: the square wave (Worked Example 2 below) is piecewise \( C^1 \) with jump discontinuities, and near each jump the partial sums overshoot the limiting value by a fixed proportion (about \( 8.9\% \) of the jump, the Gibbs phenomenon) no matter how large \( N \) is; convergence is pointwise/uniform away from the jump but never uniform on an interval containing it.

Common errors

Writing "the Fourier series converges to \( f(x) \)" at a jump discontinuity instead of to \( \frac{f(x^+)+f(x^-)}{2}\) – the series never converges to either one-sided value at a genuine jump, it converges to the average.
Believing pointwise convergence at every point implies uniform convergence on an interval; the Gibbs phenomenon shows the rate of convergence near a jump does not improve with \( N \), so convergence there is pointwise but not uniform even though every individual point converges.
Applying the theorem to a function that is only piecewise continuous but not piecewise \( C^1 \) (e.g. continuous but nowhere differentiable, à la Weierstrass) and asserting convergence at every point – the hypothesis genuinely fails and no conclusion follows from this theorem (a different, weaker theorem such as Carleson's may still apply for a.e. \( x \), but not this one for every \( x \)).
Confusing the symmetric partial sum \( S_N f(x) = \sum_{n=-N}^{N} \), which this theorem governs, with an asymmetric one-sided partial sum \( \sum_{n=0}^{2N} \); the theorem as proved via the Dirichlet kernel relies on the symmetric range, and asymmetric truncations can behave differently.
Forgetting to check periodicity when \( f \) is only given on \( [-\pi,\pi] \): if \( f(-\pi) \neq f(\pi) \), the periodic extension has a jump at \( x=\pi \) even though \( f \) "looked continuous" on the closed interval, and the theorem then gives \( \frac{f(-\pi)+f(\pi)}{2} \) there, not \( f(\pi) \).

Discussion

The theorem is usually attributed to Dirichlet (1829), who gave the first fully rigorous convergence proof for a class of functions essentially matching "piecewise monotone" (later subsumed by Jordan's 1881 extension to functions of bounded variation, using the same Dirichlet-kernel technique but replacing the derivative hypothesis with bounded variation on a neighbourhood of \( x \)). The mechanism is worth dwelling on: convergence at \( x \) is a strictly local question – this is the content of the Riemann localisation principle, visible directly in the proof, since only the behaviour of \( f \) in an arbitrarily small neighbourhood of \( x \) (through \( g_\pm \) near \( t=0 \)) controls the limit, even though the Fourier coefficients themselves are computed from an integral over the whole period.

The Gibbs phenomenon, visible in both worked examples below, is not a defect of the theorem but a genuinely separate fact about the rate and uniformity of convergence near a jump; it was observed experimentally by Michelson using a harmonic analyser machine before Gibbs (1899) and Bôcher (1906) explained it analytically, and it is a standing warning that pointwise convergence, however clean the limit, carries no information about convergence in the sup norm.

Historically the search for the right hypothesis went through several near-misses: Dirichlet's original 1829 conditions, the identification of the Riemann–Lebesgue lemma as the essential analytic tool (independently by Riemann in his 1854 Habilitationsschrift and Lebesgue after developing his integral), du Bois-Reymond's 1873 continuous-but-divergent counterexample showing that continuity is not sufficient by itself, and finally Carleson's 1966 theorem (using entirely different, far harder techniques) establishing a.e. convergence merely from \( f \in L^2 \), with Kolmogorov's 1926 example showing \( L^1 \) is not enough even for a.e. convergence. The Dirichlet–Jordan theorem sits as the accessible, classical middle ground between these two extremes.

A common misconception is that "Fourier series always converge for nice enough functions, so the hypotheses are a technicality." In fact the gap between "continuous" and "piecewise \( C^1 \)" is exactly where classical pointwise convergence can fail (du Bois-Reymond), while the gap between "piecewise \( C^1 \)" and merely "\( L^2 \)" is exactly where the far deeper Carleson theorem is required to rescue a.e. (not everywhere) convergence; the hypothesis here is doing real work, not decoration.

Worked examples

f(x) = x \ \text{on } (-\pi,\pi), \quad f(x+2\pi)=f(x) \ \text{(the sawtooth)}

Set-up: \( f \) is piecewise \( C^1 \) (smooth on \( (-\pi,\pi) \), with the periodic extension having jumps at odd multiples of \( \pi \)); compute its Fourier series to test the theorem at a jump. A

a_n = 0 \ (\text{all } n, \text{ since } f \text{ is odd}), \qquad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)\,dx = \frac{2(-1)^{n+1}}{n}

\( a_n = 0 \) because \( x\cos(nx) \) is odd and the integral of an odd function over a symmetric interval vanishes; \( b_n \) is computed by integration by parts, \( \int x\sin(nx)\,dx = -\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2} \), evaluated on \( [-\pi,\pi] \). A

S_N f(x) = \sum_{n=1}^{N} \frac{2(-1)^{n+1}}{n}\sin(nx)

Substitute the coefficients from Step 2 into the definition of \( S_N f \). A

x = \pi : \quad f(\pi^+) = -\pi, \ \ f(\pi^-) = \pi \ \ \Rightarrow \ \ \frac{f(\pi^+)+f(\pi^-)}{2} = 0

Since \( f \) is \( 2\pi \)-periodic and equals \( x \) on \( (-\pi,\pi) \), approaching \( x=\pi \) from below gives values near \( \pi \), while approaching from above lands in the next period near \( -\pi \); this is exactly the jump configuration the theorem addresses. A

S_N f(\pi) = \sum_{n=1}^{N} \frac{2(-1)^{n+1}}{n}\sin(n\pi) = 0 \ \text{for every } N

\( \sin(n\pi) = 0 \) for every integer \( n \), so every partial sum is identically \( 0 \) at \( x=\pi \); the theorem's conclusion \( S_N f(\pi) \to 0 = \frac{f(\pi^+)+f(\pi^-)}{2} \) holds trivially and exactly, confirming Step 4. A

S_N f(\pi) = 0 = \frac{f(\pi^+)+f(\pi^-)}{2} \ \text{for all } N, \ \text{consistent with the theorem}

Reading. At the jump, the series sits exactly at the midpoint for every partial sum, the cleanest possible illustration of the theorem's conclusion.

f(x) = \begin{cases} 1 & 0 \lt x \lt \pi \\ -1 & -\pi \lt x \lt 0 \end{cases}, \quad f(0)=f(\pi)=0, \quad f(x+2\pi)=f(x)

Set-up: the square wave, piecewise \( C^1 \) with jumps at every integer multiple of \( \pi \); the goal is to evaluate the series at the continuity point \( x=\pi/2 \) to extract a numerical identity. A

a_n = 0 \ (\text{odd } f), \qquad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)\sin(nt)\,dt = \frac{2}{\pi}\int_0^\pi \sin(nt)\,dt = \frac{2\big(1-(-1)^n\big)}{n\pi} = \begin{cases} \frac{4}{n\pi} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}

Oddness kills \( a_n \) as before; \( b_n \) uses \( f(t)\sin(nt) \) even (product of two odd functions) to double the integral over \( [0,\pi] \), then evaluates \( \int_0^\pi \sin(nt)\,dt = \frac{1-\cos(n\pi)}{n} = \frac{1-(-1)^n}{n} \). A

S_N f(x) = \frac{4}{\pi}\sum_{\substack{n=1 \\ n \text{ odd}}}^{N} \frac{\sin(nx)}{n}

Substitute the nonzero coefficients from Step 2 into the definition of \( S_N f \). A

x = \frac{\pi}{2} \ \text{is a point of continuity of } f, \ \text{with } f\!\left(\frac{\pi}{2}\right) = 1, \ \text{ and } f \text{ is piecewise } C^1 \text{ at } x=\tfrac{\pi}{2}

\( f \equiv 1 \) on the open interval \( (0,\pi) \ni \pi/2 \), so \( f \) is constant, hence \( C^1 \), in a neighbourhood of \( \pi/2 \); both hypotheses of the theorem hold there with \( f(x^+)=f(x^-)=f(x)=1 \). A

\sin\!\left(\frac{n\pi}{2}\right) = \begin{cases} (-1)^{(n-1)/2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases} \ \Rightarrow \ S_N f\!\left(\tfrac{\pi}{2}\right) = \frac{4}{\pi}\left(1 - \frac13 + \frac15 - \frac17 + \cdots \right)

Direct evaluation of \( \sin(n\pi/2) \) for \( n=1,3,5,7,\dots \) gives the alternating pattern \( 1,-1,1,-1,\dots \); substituting into the sum of Step 3 at \( x=\pi/2 \) produces the alternating harmonic-type series over odd denominators. B

\text{By the theorem, } \lim_{N\to\infty} S_N f\!\left(\tfrac{\pi}{2}\right) = f\!\left(\tfrac{\pi}{2}\right) = 1

Direct application of the theorem's conclusion at the continuity point identified in Step 4: since \( f(x^+)=f(x^-)=1 \), the average is \( 1 \). A

1 - \frac13 + \frac15 - \frac17 + \cdots = \frac{\pi}{4}

Reading. Setting the limit from Step 5 equal to \( 1 \) from Step 6 and solving gives the Leibniz series for \( \pi/4 \) – a genuine numerical consequence of the convergence theorem applied at a continuity point, not merely at a jump.

Problems

Let \( f(x) = |x| \) on \( [-\pi,\pi] \), extended \( 2\pi \)-periodically. Verify the hypotheses of the theorem hold at every point, compute the Fourier series, and use the theorem at \( x=0 \) to evaluate \( \sum_{k=1}^\infty \frac{1}{(2k-1)^2} \).
Solution
\( f \) is continuous everywhere (the periodic extension of \( |x| \) has no jumps, since \( f(-\pi)=f(\pi)=\pi \)) and piecewise \( C^1 \) (smooth away from \( x=0, \pm\pi \), with one-sided derivatives \( \mp1 \) existing at those corners), so the theorem applies at every \( x \) and gives convergence to \( f(x) \) everywhere. Since \( f \) is even, \( b_n=0\); \( a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}|x|\,dx = \pi \), and for \( n\geq1 \), \( a_n = \frac{2}{\pi}\int_0^\pi x\cos(nx)\,dx = \frac{2}{\pi}\left[\frac{x\sin(nx)}{n}+\frac{\cos(nx)}{n^2}\right]_0^\pi = \frac{2\big((-1)^n-1\big)}{\pi n^2}\), which is \( -\frac{4}{\pi n^2} \) for odd \( n \) and \( 0 \) for even \( n \). So \( f(x) = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=1}^\infty \frac{\cos((2k-1)x)}{(2k-1)^2}\). At \( x=0 \), \( f(0)=0 \) and the theorem gives \( 0 = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=1}^\infty \frac{1}{(2k-1)^2}\), so \( \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}\).
Explain precisely why this theorem cannot be invoked to conclude pointwise convergence of the Fourier series of the Weierstrass function \( W(x) = \sum_{k=0}^\infty a^k \cos(b^k \pi x) \) (with \( 0\lt a\lt1 \), \( ab \gt 1\)) at any point, and state what, if anything, can still be said.
Solution
\( W \) is continuous and periodic but is nowhere differentiable, so at every point \( x \) it fails to have even a one-sided derivative; the hypothesis of the theorem fails everywhere, so the theorem gives no information at any point (this is not the same as proving divergence). Since \( W \) is continuous, Fejér's theorem still guarantees the Cesàro (arithmetic mean) partial sums converge uniformly to \( W \); and being continuous and hence bounded and measurable, \( W \in L^2(\mathbb{T}) \), so Carleson's theorem guarantees the ordinary partial sums \( S_N W(x) \) converge to \( W(x) \) for almost every \( x \), though this is a vastly harder theorem than the one proved here and does not pin down convergence at any specific named point.
Prove the closed-form identity used in Step 1 of the proof: for \( t \notin 2\pi\mathbb{Z} \), \( \sum_{n=-N}^{N} e^{int} = \frac{\sin\left((N+\tfrac12)t\right)}{\sin(t/2)} \).
Solution
Let \( z=e^{it}\), \( z\neq1\) since \( t\notin2\pi\mathbb{Z}\). Then \( \sum_{n=-N}^N z^n = z^{-N}\sum_{k=0}^{2N} z^k = z^{-N}\cdot\frac{z^{2N+1}-1}{z-1}\) by the finite geometric series formula. Multiply numerator and denominator by \( z^{-1/2}=e^{-it/2}\): \(\frac{z^{2N+1}-1}{z-1}\cdot z^{-N} = \frac{z^{N+1}-z^{-N}}{1}\cdot z^{-N}\)... more directly, \( z^{-N}(z^{2N+1}-1) = z^{N+1}-z^{-N}\), so the sum equals \(\frac{z^{N+1}-z^{-N}}{z-1}\). Multiply top and bottom by \( z^{-1/2}\): \(\frac{z^{N+1/2}-z^{-N-1/2}}{z^{1/2}-z^{-1/2}} = \frac{e^{i(N+1/2)t}-e^{-i(N+1/2)t}}{e^{it/2}-e^{-it/2}} = \frac{2i\sin\left((N+\tfrac12)t\right)}{2i\sin(t/2)} = \frac{\sin\left((N+\tfrac12)t\right)}{\sin(t/2)}\), using \( e^{i\theta}-e^{-i\theta}=2i\sin\theta\) in both numerator and denominator.
Let \( f \) be the \( 2\pi \)-periodic function with \( f(x) = \pi - x \) for \( 0 \lt x \lt 2\pi \) (extended periodically, undefined value assigned as the midpoint at multiples of \( 2\pi \)). State the value the Fourier series converges to at \( x=0 \), and verify it against the known Fourier series \( f(x) = 2\sum_{n=1}^\infty \frac{\sin(nx)}{n} \).
Solution
The periodic extension has \( f(0^-) = \lim_{x\to2\pi^-}(\pi-x) = -\pi\) and \( f(0^+) = \lim_{x\to0^+}(\pi-x)=\pi\), so \( x=0\) is a jump and the theorem predicts \( \lim_N S_Nf(0) = \frac{\pi+(-\pi)}{2}=0\). Checking directly: \( S_Nf(0) = 2\sum_{n=1}^N \frac{\sin(0)}{n} = 0\) for every \( N\), since \( \sin(0)=0\); the partial sums are identically \( 0\), matching the predicted limit exactly, just as in Worked Example 1.
Using the theorem, justify why the Fourier series of a piecewise-\( C^1 \) periodic function \( f \) that is continuous everywhere and satisfies \( f(x)=f(-x)\) (even) must converge to \( f(x)\) at \( x=0\) and \( x=\pi\) in particular, and use this together with the coefficient formula to show that if additionally \( a_n \geq 0\) for all \( n\) and \( \sum a_n \lt \infty\), then \( f(x) = \frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(nx)\) converges absolutely and uniformly, not just pointwise.
Solution
Continuity of \( f \) at every point, combined with piecewise-\(C^1\), means the theorem applies at every \( x\), in particular \( x=0\) and \( x=\pi\), giving \( S_Nf(0)\to f(0)\) and \( S_Nf(\pi)\to f(\pi)\) — this only re-confirms ordinary pointwise convergence at those two points, nothing stronger yet. For the second part: if \( a_n\geq0\) and \( \sum_{n=1}^\infty a_n\lt\infty\), then since \( |a_n\cos(nx)|\leq a_n\) for every \( x\), the Weierstrass M-test with \( M_n=a_n\) shows \( \sum a_n\cos(nx)\) converges absolutely and uniformly on \( \mathbb{R}\) to some continuous function \( g(x)\); pointwise, the theorem already forces \( g(x)=\lim_N S_Nf(x) = f(x)\) at every point of continuity, i.e. everywhere by hypothesis, so \( g=f\) and the convergence upgrades from the theorem's pointwise statement to full uniform convergence, precisely because absolute summability of the coefficients was available as extra information the theorem alone does not provide.