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The DC component of a periodic signal is its constant, zero-frequency part. For an integrable signal $f$ of period $T>0$, it is the time averageThe result is independent of $t_0$ when a whole period is used. In a complex Fourier series with coefficients $c_n=T^{-1}{\int }_0^{T}f(t)e^{-2\pi int/T}dt$, the DC component is $c_0$. In the convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos (2\pi nt/T)+b_n\sin (2\pi nt/T))$, it is $a_0/2$, not $a_0$. The constant character and Fourier coefficient at zero are part of the compact-group Fourier series description.^[ref]

Write $f=\mu +g$, so $⟨g⟩=0$. For a real signal with finite mean square,because the cross term $2\mu ⟨g⟩$ vanishes. Thus total root mean square combines the DC component and the varying part quadratically. A meter that removes the DC component and a meter reporting total root mean square need not give the same number.

For an original rectangular waveform, take $f=8$ for one quarter of each period and $f=-2$ for the rest. Then $\mu =8/4-2(3/4)=0.5$, whereas $⟨f^2⟩=19$. Its varying part has root mean square $\sqrt{19-0.5^2}=\sqrt{18.75}\approx 4.330$. The offset is small even though the instantaneous values are not.

An observation window that covers a noninteger number of periods can give a different sample average. For example, a zero-mean sinusoid observed over only part of a cycle need not have a zero window average. Finite-window averaging therefore estimates the periodic DC component only under an appropriate sampling model; a nonzero finite record mean can also reflect drift or an incomplete cycle.

The Du Bois-Reymond theorem says that continuity of a periodic function does not guarantee convergence of its ordinary Fourier series at every point. At any prescribed circle point $x_0$, there is even a real continuous $f$ with unbounded Fourier partial sums $S_{N}f(x_0)$.

For $x_0=0$, view $T_N:f\mapsto S_{N}f(0)$ as bounded linear functionals on the real Banach space $C(\mathbb{T})$ with its supremum norm. The Dirichlet kernel harmonic lower bound givesIf every continuous input had bounded values under all $T_N$, the Uniform boundedness principle would bound these operator norms, a contradiction. Translation moves the exceptional point to any specified $x_0$. The generic unbounded Fourier sums at a fixed point strengthens this existence proof to a category statement.^[ref]

This does not contradict mean-square convergence: a continuous function is square-integrable, and its series still converges in $L^2$. Nor does it obstruct Fejér's theorem, which averages partial sums and gives uniform convergence for every continuous periodic input. The Fejér kernel is positive and its sums are a uniform-norm contraction, but the sums themselves need not be nonnegative: for $f=-1$, every Fejér sum equals $-1$.

Use $N\ge 1$ terms in the arithmetic mean. The Fejér kernel isHere $D_j$ is the Dirichlet kernel. Counting pairs with a fixed index difference proves the middle coefficient formula. The finite geometric series givesThe squared-modulus form proves nonnegativity, and the constant Fourier coefficient gives $(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{F}_N=1$. Thus $K_N=F_N/(2\pi )$ is the unit-mass convolution density. Another common half-kernel convention uses $H_N=F_N/2$ with convolution prefactor $1/\pi$; these normalizations must be distinguished.

For each fixed $\delta >0$, $F_N(t)\le [N{\sin }^2(\delta /2){]}^{-1}$ when $\delta \le |t|\le \pi$, so the mass away from zero tends to zero. Positivity, total mass and this concentration make the kernels an approximate identity, explaining Fejér's theorem. In this convention the degree is $N-1$, while $N$ is the number of averaged partial sums.^[ref]

Take the Gaussian function $f_a(x)=e^{-\pi a{x}^2}$ with $a>0$. Every derivative is a polynomial times the same Gaussian function, so $f_a$ is a Schwartz function. The Fourier transform of a Gaussian gives, in the cycles convention,Thus the Poisson summation formula gives the exact identityThe Gaussian Fourier transform is given in scaled form in Exercise 41 of Terence Tao's 245C, Notes 2: The Fourier transform.^[ref] The identity is also the imaginary-axis transformation of the Jacobi theta function. At $a=1$ the Gaussian function is its own Fourier transform; at reciprocal values of $a$ its spatial and spectral widths exchange.

For a short derivation of the Gaussian Fourier transform, write $H(\xi )={\hat{f}}_a(\xi )$. Differentiation under the integral and integration by parts, using $f_a^{\prime }(x)=-2\pi ax{f}_a(x)$, giveThe Gaussian integral gives $H(0)=a^{-1/2}$, so this ordinary differential equation yields the claimed formula. The decay of the Gaussian function justifies these steps and removes the boundary term.^[ref]

For $a=0.04$, the physical Gaussian function has width parameter $1/\sqrt{a}=5$, while its Fourier transform has width parameter $\sqrt{a}=0.2$. Many physical samples matter, but the dual zero sample already givesFor $a=25$, use the physical side instead:These are bounds on the full positive tails, obtained below, rather than estimates based only on the first omitted term. Ordinary floating-point arithmetic rounds these sums to $5$ and $1$; the exact real numbers are slightly larger. A Gaussian sum truncation bound lets us choose the cheaper side and certify its truncation error before evaluating it.

For a Fourier multiplier $m\in L^{\infty }({\mathbb{R}}^d)$, define $T_{m}f={\mathcal{F}}^{-1}(m\mathcal{F}f)$ using a unitary Fourier transform on L2 space. Its operator norm is exactly the essential supremum of $|m|$:The unitary normalization matters when writing the intermediate integrals. For example, $\hat{f}(\xi )=\int f(x)e^{-2\pi ix\cdot \xi }dx$ has this normalization. The Plancherel theorem then turns the bound into multiplication of functions:The required extension of the Fourier transform to all of L2 space is the unitary extension in the Plancherel theorem.^[ref]

To prove sharpness, write $M=\Vert m{\Vert }_{\infty }>0$. For every $0<\epsilon <M$, the measurable set where $|m|>M-\epsilon$ has positive Lebesgue measure. Choose a subset $E$ of finite positive Lebesgue measure and prescribe $\hat{f}=1_E/\sqrt{|E|}$, using an indicator function. The inverse Fourier transform gives $\Vert f{\Vert }_2=1$ and $\Vert T_{m}f{\Vert }_2\ge M-\epsilon$. Letting $\epsilon$ tend to zero proves the equality of operator norms. The case $M=0$ is immediate.

For example, $m=1_{[-1,1]}$ is an ideal low-pass filter: it is an orthogonal projection onto the signals with Fourier transform supported in that interval, and has operator norm one. In contrast, $m(\xi )=e^{-{\xi }^2}$ also has operator norm one on $L^2(\mathbb{R})$, but no nonzero signal attains equality: $|m|<1$ except on a set of Lebesgue measure zero. Changing a Fourier multiplier at one frequency cannot change this operator norm.

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.

For a complex weakly stationary process with finite second moment and continuous uncentered autocorrelation $\Gamma (\tau )=\mathbb{E}[\overline{X_t}{X}_{t+\tau }]$, there is a unique finite positive spectral measure of a stationary random field. With the optical sign convention, write it asThe sign reverses the frequency coordinate relative to the alternative positive-exponential convention. To see the positivity behind the theorem, expand $\mathbb{E}|{\sum }_j{c}_j{X}_{t_j}{|}^2\ge 0$. This makes the correlation a positive-definite function, so Bochner's theorem supplies the measure.^[ref]

An ordinary power spectral density $S$ exists when $dF=S(\omega )d\omega$. If $\Gamma$ is integrable, $S(\omega )=(2\pi {)}^{-1}\int \Gamma (\tau )e^{i\omega \tau }d\tau$. Stationarity alone does not guarantee such a density. For $X_t=Z$ at every time and $\mathbb{E}|Z{|}^2=1$, the correlation is one and $F={\delta }_0$.

The normalized Lorentzian line shape $S(\omega )=\kappa /[\pi ((\omega -{\omega }_0{)}^2+{\kappa }^2)]$ gives $\Gamma (\tau )=e^{-\kappa |\tau |}{e}^{-i{\omega }_0\tau }$. Its width therefore describes decay of temporal coherence. A persistent spectral line instead contributes an atom, which cannot be treated as an ordinary integrable density.