Three Different Ways a Sequence Can Converge
For Fourier partial sums, convergence in \(L^2\) means the integral of the squared error tends to zero. It does not mean that the error tends to zero at every point, nor does it mean that the convergence is uniform. These distinctions matter because an \(L^2\) function is identified up to changes on a set of measure zero, whereas pointwise values refer to a particular representative of that equivalence class.
Use the circle \(\mathbb T=\mathbb R/\mathbb Z\), represented by \([0,1)\), with its completed Lebesgue sigma-algebra and normalized Lebesgue measure \(\mu\), so \(\mu(\mathbb T)=1\). Functions in this tutorial are complex-valued and measurable unless stated otherwise. For \(f\in L^2(\mathbb T,\mu)\), the Fourier coefficient at \(k\in\mathbb Z\) and the partial sum of order \(N\) are
The coefficient integral is defined: since \(\mu(\mathbb T)=1\), the Cauchy–Schwarz inequality gives \(\int_{\mathbb T}|f|\,d\mu\le\|f\|_2\), and \(|e^{-2\pi i kx}|=1\). The earlier results on completeness of the trigonometric system and orthogonal projections give the \(L^2\) Fourier convergence theorem: for every \(f\in L^2(\mathbb T,\mu)\), \(\|S_Nf-f\|_2\to0\). This is convergence in norm, not a statement about pointwise values.
What \(L^2\) Convergence Does Imply
On this finite measure space, \(L^2\) convergence implies convergence in measure. Convergence in measure means that for every \(\varepsilon>0\), the measure of the strict-error set \(\{|f_n-f|>\varepsilon\}\) tends to zero. This conclusion is weaker than pointwise convergence of the whole sequence, but it gives a useful quantitative control at each fixed threshold.
Proof. Fix \(\varepsilon>0\), and put \(E_n=\{x:|f_n(x)-f(x)|>\varepsilon\}\). On \(E_n\), \(|f_n-f|^2>\varepsilon^2\), so monotonicity of the integral gives \(\varepsilon^2\mu(E_n)\le\int_{E_n}|f_n-f|^2\,d\mu\le\|f_n-f\|_2^2\). Therefore \(0\le\mu(E_n)\le\varepsilon^{-2}\|f_n-f\|_2^2\to0\), as required. \(\square\)
A second useful fact is that an \(L^2\)-convergent sequence has a subsequence converging almost everywhere. This does not say that the original sequence converges almost everywhere.
Proof. Since \(\|f_n-f\|_2\to0\), choose increasing indices \(n_j\) such that \(\|f_{n_j}-f\|_2\le2^{-2j}\) for every \(j\ge1\). Let \(E_j=\{|f_{n_j}-f|>2^{-j}\}\). Applying the estimate in the preceding proof with threshold \(2^{-j}\) gives \(\mu(E_j)\le 2^{2j}\|f_{n_j}-f\|_2^2\le2^{2j}2^{-4j}=2^{-2j}\). Thus \(\sum_{j=1}^{\infty}\mu(E_j)\le\sum_{j=1}^{\infty}2^{-2j}=1/3<\infty\). By countable subadditivity, \(\mu(\bigcup_{j\ge r}E_j)\le\sum_{j\ge r}\mu(E_j)\to0\) as \(r\to\infty\). The sets \(\bigcup_{j\ge r}E_j\) decrease with \(r\), and the first has finite measure, so continuity from above gives \(\mu(\limsup_j E_j)=0\). Outside this null set, each point belongs to only finitely many \(E_j\). At every such point, \(|f_{n_j}-f|\le2^{-j}\) for all sufficiently large \(j\), and hence \(f_{n_j}\to f\). \(\square\)
For Fourier partial sums, the earlier \(L^2\) convergence theorem therefore also guarantees a subsequence converging almost everywhere to \(f\). It does not, by itself, establish pointwise convergence of the full sequence of partial sums. The separate Carleson theorem asserts almost-everywhere convergence of Fourier partial sums for \(L^2\) functions; this is a deeper pointwise theorem, not a consequence of the norm-convergence statement proved by orthogonal projection.
Worked Examples: Norm, Pointwise Values, and Fourier Sums
Worked Example: An \(L^2\)-Convergent Sequence with No Pointwise Limit
On \((\mathbb T,\mu)\), construct a sequence converging to zero in \(L^2\) that fails to converge pointwise anywhere. Partition \([0,1)\), at each level \(k\ge1\), into the half-open intervals \(I_{k,j}=[j2^{-k},(j+1)2^{-k})\), where \(j=0,\ldots,2^k-1\). Enumerate their indicators level by level: the \(k\)-th block consists of \(\chi_{I_{k,0}},\ldots,\chi_{I_{k,2^k-1}}\). Each function is measurable, and for every \(j\),
Thus every function in block \(k\) has norm \(2^{-k/2}\). The block number tends to infinity as the sequence index tends to infinity, so these norms tend to zero and the sequence converges to zero in \(L^2\). But each \(x\in[0,1)\) belongs to exactly one interval at every level \(k\), so its function values include a \(1\) in every block. For \(k\ge2\), the block also contains intervals not containing \(x\), so the values include a \(0\) in every such block. The values at \(x\) therefore do not converge. Check: the level-\(k\) norm is \(2^{-k/2}\), which tends to \(0\), while the pointwise values have infinitely many \(1\)'s and infinitely many \(0\)'s. This is an \(L^2\) sequence, not a sequence of Fourier partial sums.
Worked Example: Pointwise Convergence Without \(L^2\) Convergence
On \((\mathbb T,\mu)\), let \(f_n=n\chi_{(0,1/n)}\). For each fixed \(x>0\), if \(n>1/x\), then \(x\notin(0,1/n)\), so \(f_n(x)=0\) thereafter. At \(x=0\), the interval is open and \(f_n(0)=0\) for every \(n\). Hence \(f_n(x)\to0\) at every point. However,
Thus \(\|f_n\|_2=\sqrt n\), which does not tend to zero. The pointwise limit is the zero function, whose \(L^2\) norm is \(0\), so pointwise convergence here is not \(L^2\) convergence. Check: for \(n=4\), the support has measure \(1/4\), and the squared norm is \(4^2(1/4)=4\), agreeing with the formula \(n=4\).
Worked Example: Changing One Value Leaves Fourier Coefficients Unchanged
Define \(f(x)=1\) for every \(x\in\mathbb T\), and define \(g(x)=1\) for \(x\ne0\) but \(g(0)=5\). These bounded measurable functions differ only on \(\{0\}\), whose measure is zero. For every \(k\in\mathbb Z\), the integral of \((g-f)e^{-2\pi ikx}\) is zero because its integrand vanishes outside that null set. Therefore \(\widehat g(k)=\widehat f(k)\). Directly, the constant function has \(\widehat f(0)=\int_{\mathbb T}1\,d\mu=1\), and for \(k\ne0\),
Consequently \(S_Ng=S_Nf=1\) everywhere for every \(N\ge0\). The partial sums equal \(g\) at every \(x\ne0\), but at \(x=0\) they equal \(1\), not \(g(0)=5\). This does not contradict \(L^2\) convergence: \(f=g\) \(\mu\)-a.e., so they represent the same element of \(L^2(\mathbb T,\mu)\). Check: the norm of their difference is \(\|g-f\|_2^2=\int_{\{0\}}16\,d\mu=16\cdot0=0\), as required.
Uniform Convergence and Common Misreadings
On \((\mathbb T,\mu)\), uniform convergence to a representative \(f\) implies \(L^2\) convergence to \(f\): if \(\sup_{x\in\mathbb T}|f_n(x)-f(x)|\le c_n\) and \(c_n\to0\), then
The converse fails for general measurable functions. For example, on this same circle let \(h_n=\chi_{(0,1/n)}\). Then \(\|h_n\|_2^2=1/n\to0\), but \(\sup_{x\in\mathbb T}|h_n(x)|=1\) for every \(n\), so convergence to zero is not uniform. This example concerns a sequence of functions, not Fourier partial sums. Also distinguish a uniform bound on a chosen representative from an essential-supremum bound: the latter ignores null sets and does not assert uniform convergence at every point.
A frequent error is to infer pointwise convergence from a small \(L^2\) error, or to say that Fourier partial sums must converge at every point because they converge in \(L^2\). The interval-indicator sequence above disproves the first inference for general sequences; it does not disprove the separate Carleson theorem for Fourier partial sums. Another error is to treat a Fourier series as determining the value of a function at every point. Fourier coefficients depend only on the \(L^2\) equivalence class, so they cannot distinguish representatives that differ on a null set. Any pointwise claim must specify a representative and invoke a pointwise convergence result, not just norm convergence.
Check Your Understanding
For each question, distinguish statements about a general \(L^2\) sequence from statements specifically about Fourier partial sums.
- Why does \(\|f_n-f\|_2\to0\) imply \(\mu(\{|f_n-f|>\varepsilon\})\to0\) for a fixed \(\varepsilon>0\)?
- What feature of the level-by-level interval sequence makes it fail to converge pointwise at each \(x\), despite convergence in \(L^2\)?
- On \(\mathbb T\), does uniform convergence imply \(L^2\) convergence? State the finite-measure estimate that answers this.
- Why do two functions equal \(\mu\)-a.e. have identical Fourier coefficients at every integer frequency?
- Does the \(L^2\) Fourier convergence theorem alone imply pointwise convergence of the full sequence of partial sums? Name the separate theorem relevant to that pointwise assertion.