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If the constant symmetric matrix $A$ satisfies $A\xi \cdot \xi \ge \theta |\xi {|}^2$, then every smooth function $u$ with compact support satisfiesIndeed, the Fourier transform of a derivative and the Plancherel theorem turn the squared norms into integrals with Fourier multipliers $|\xi {|}^4$ and $(A\xi \cdot \xi {)}^2$.

The Fourier inversion theorem makes decomposition into frequencies reversible: a complete Fourier transform, including its wave phases, determines the original function up to equality almost everywhere. With the angular-frequency conventionif $f$ and $\hat{f}$ belong to the L1 space, thenalmost everywhere. The right side is a continuous representative of $f$. If $f$ itself is continuous, equality holds everywhere, as explained by the continuous version of Fourier inversion. The integrability hypotheses ensure that both displayed integrals are ordinary absolutely convergent integrals. For arbitrary elements of the L2 space, the Plancherel theorem instead supplies inversion in the L2 space norm.

Each factor $e^{ix\cdot \xi }$ is a Fourier mode; the inverse integral adds them with the complex weights $\hat{f}(\xi )/(2\pi {)}^n$. The normalization is part of the recipe. In one dimension, writing ordinary frequency as $\nu =\xi /(2\pi )$ gives the equivalent pairwhere $F(\nu )=\hat{f}(2\pi \nu )$. Reversing both exponential signs gives another equivalent convention, but the forward and inverse formulas must use opposite signs consistently.

For example, the Fourier transform of a Gaussian givesFourier inversion says that a weighted continuum of oscillations reconstructs this smooth Gaussian function. Narrowing the original Gaussian function by increasing $a$ broadens its frequency spectrum. This is a concrete demonstration that a sharply localized shape needs a wider range of frequencies.

Inversion is what makes a frequency calculation useful: transform a function, solve the simpler problem there, then return to the original variables. A brief proof outline is to multiply $\hat{f}(\xi )$ by $e^{-ϵ|\xi {|}^2}$, $ϵ>0$. The inverse integral becomes convolution with a normalized Gaussian function, an approximate identity. As $ϵ$ tends to zero, dominated convergence gives the unregularized inverse integral, while the approximate identity recovers $f$ almost everywhere. Joseph Fourier's Théorie analytique de la chaleur (1822), Chapter IX, article 417, contains a cosine double-integral representation antecedent to this modern formula; the hypotheses above are the modern statement.^[ref]

A Fourier multiplier operator $T_m$ is defined by $\widehat{T_{m}f}(\xi )=m(\xi )\hat{f}(\xi )$. On $L^2$, a bounded symbol gives a bounded operator by the Plancherel theorem, while an unbounded symbol defines a closed linear operator on the functions for which $m\hat{f}$ remains square-integrable.

Functions with pairwise disjoint, or uniformly finitely overlapping, Fourier supports are almost orthogonal in $L^2$ by the Plancherel theorem.

Consider a uniform plane wave of unit complex amplitude transmitted by an ideal optical slit of full width $D$. Treat the field as invariant along the optical slit height: this is one-dimensional transverse Fresnel propagation, so the prefactor is $1/\sqrt{i\lambda z}$, rather than the two-dimensional prefactor $1/(i\lambda z)$. Removing the carrier $e^{ikz}$ givesHere $I_{\mathrm{inc}}$ is incident irradiance. This infinitely long slit model describes a transverse cut away from the ends of a sufficiently long physical slit; its finite conserved quantity is transmitted power per unit optical slit height.

Put $t=\xi /D$ and $q=xD/(\lambda z)$. Substitution reduces the integral toThis form separates the pattern's shape from its physical scale. Reflection $t\mapsto -t$ proves $A(-q,N_D)=A(q,N_D)$, so the light intensity is an even function of $x$. At $N_D=0$, direct integration gives $A(q,0)=\sin (\pi q)/(\pi q)$, with limit one at $q=0$. Hence the Fraunhofer diffraction comparison is the squared sinc function. In this paragraph the unnormalized sinc function takes argument $\pi q$.

For an original example, choose wavelength $\lambda =633\mathrm{nm}$ and full width $D=0.400\mathrm{mm}$. At $z=0.0200\mathrm{m}$, $N_D\simeq 12.6382$, $N_a\simeq 3.15956$ and the edge-to-centre wave phase change is $9.92604$ radians. At $z=2.00\mathrm{m}$, $N_D\simeq 0.126382$, $N_a\simeq 0.0315956$ and that wave phase change is only $0.0992604$ radians. Even the nearer screen is fifty slit widths away; that geometric comparison alone misses its strong Fresnel diffraction.

These are numerical predictions from the displayed scalar integral, using the composite trapezoidal rule, rather than measured profiles. Each curve is divided by its own peak light intensity; the two horizontal scales differ. At 2 cm the computed central irradiance is about $1.14812I_{\mathrm{inc}}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\mathrm{inc}}$ and is the peak. The dashed Fraunhofer diffraction curve has its own central normalization; its first zeros are at $x=\pm \lambda z/D=\pm 3.165\mathrm{mm}$. The finite-distance Fresnel diffraction curve remains slightly nonzero there.

Normalization hides the brightness change, but not the change in shape. The Fourier transform propagation multiplier has unit complex modulus, so Plancherel theorem givesThis is conservation of transmitted power per unit height in the ideal scalar model. An integral over only the displayed screen interval misses the tails; a peak-normalized profile is not a unit-integral probability density. Spreading and decreasing brightness together preserve the transmitted power.

The half-plane Fourier projection is $(I-i{H}_v)/2$, where $H_v$ is the directional Hilbert transform. A disk of radius $R$, modulated to put a boundary point at the origin, has symbol $1_{\{|\xi +Rv|\le R\}}$, which converges almost everywhere to this half-plane symbol. The Plancherel theorem makes this an $L^2$ multiplier limit.

For $\mathcal{H}f(x)={\pi }^{-1}\mathrm{pv}\int f(y)/(x-y)dy$ and $\hat{f}(\xi )=\int e^{-ix\xi }f(x)dx$, the Fourier transform multiplier is $-i\mathrm{sgn}\xi$. The Heaviside function transform and the double-transform reflection identity give the sign. Plancherel theorem proves the $L^2$ isometry, and multiplication by $i\xi$ proves commutation with differentiation. The map need not preserve the Schwartz space.

If a finite measure on ${\mathbb{R}}^n$ has square-integrable Fourier transform, the Plancherel theorem gives an $L^2$ inverse transform $g$. Pairing with every Schwartz function and using Fourier inversion identifies the measure with $gdx$. This establishes absolute continuity, rather than presuming it. For a positive measure, $g\ge 0$ and $g\in L^1$ as well. Multiplication by $f\in L^{\infty }(\mu )$ yields $\Vert \widehat{fd\mu }{\Vert }_2\le \Vert \hat{\mu }{\Vert }_2\Vert f{\Vert }_{L^{\infty }(\mu )}$.

Use $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$ and let $P_j$ be the orthogonal projection onto the closed span of the orthonormal translates and dilates of a scaling function. If $\hat{f}$ is supported in $[-R,R]$ and $R<\pi 2^j$, thenIndeed, the Plancherel theorem writes the coefficient against ${\phi }_{j,k}$ as $2^{-j/2}(2\pi {)}^{-1}\int \hat{f}(\xi )\overline{\hat{\phi }(2^{-j}\xi )}{e}^{ik{2}^{-j}\xi }d\xi$. With $\xi =2^j\theta$, these are $2^{j/2}$ times the Fourier series coefficients of $\hat{f}(2^j\theta )\overline{\hat{\phi }(\theta )}$ on $[-\pi ,\pi ]$. The Parseval identity proves the formula. It shows that continuity and unit modulus at zero imply density of the refinement spaces, and conversely that density forces this unit modulus when the Fourier transform is continuous at zero.

The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

With the Fourier convention $\hat{f}(t)=\int f(x)e^{-itx}dx$, translation by $k$ multiplies the transform by $e^{ikt}$. The Plancherel theorem gives the displayed correlation identity for every square-integrable function, by extension from integrable square-integrable functions. Thus distinct integer translates are orthogonal exactly when the nonzero integer Fourier integrals of $|\hat{f}{|}^2$ vanish. The products in both integrals are integrable by the Cauchy-Schwarz inequality.

With $\hat{\varphi }(\xi )=\int \varphi (x)e^{-ix\xi }dx$ in the Plancherel theorem sense, the integer translates of $\varphi \in L^2(\mathbb{R})$ are orthonormal exactly whenThe Fourier coefficients of the periodized energy are the translate inner products. Uniqueness of Fourier coefficients in L1 therefore proves both directions. No absolute integrability assumption on $\varphi$ is required.

Set $\tilde{\varphi }(x)=\overline{\varphi (-x)}$. This is a Schwartz function and its Fourier transform is $\hat{\tilde{\varphi }}(\xi )=\overline{\hat{\varphi }(\xi )}$. The convolution theorem givesThe convolution of two Schwartz functions is again a Schwartz function: differentiating under the integral and expanding $x^m=(y+(x-y){)}^m$ bounds every weighted derivative by products of integrable and bounded weighted derivatives. Evaluate Fourier inversion for this convolution at zero:The Schwartz space is dense in the L2 space. For $f\in L^2(\mathbb{R})$ choose Schwartz functions ${\varphi }_j\to f$ in the L2 norm. The identity makes ${\hat{\varphi }}_j$ a Cauchy sequence in the L2 space, and its limit defines $\mathcal{F}f$, independently of the approximation. It agrees with the distributional Fourier transform, because Cauchy-Schwarz inequality makes convergence in the L2 norm imply convergence when paired with every Schwartz function. For $f\in L^1\cap L^2$ it therefore agrees almost everywhere with the ordinary Fourier transform.

The extension is onto: its image is closed by the norm identity and contains the Schwartz space, since Fourier inversion maps that space to itself. This proves Plancherel theorem:Its inverse is the extension of the inverse Fourier transform, with limits understood in the L2 norm.

Use $\hat{f}(t)=\int f(x)e^{-itx}dx$ and its extension to square-integrable functions. For $T_{k}f(x)=f(x+k)$, changing variables gives $\widehat{T_{k}f}(t)=e^{ikt}\hat{f}(t)$. The Plancherel theorem therefore givesFor an arbitrary square-integrable $f$, the translation identity follows by approximation by integrable square-integrable functions: translations are isometries in $L^2$ and the Fourier transform is continuous in the Plancherel norm. The Cauchy-Schwarz inequality makes the product on the left integrable, and $|\hat{f}{|}^2$ is integrable on the right.

Put $m=l-k$. Every unequal integer pair gives a nonzero integer $m$, and every nonzero integer arises, for example with $k=0,l=m$. The identity thus proves both directions of the requested equivalence. It is the orthogonality of integer translates criterion. A different Fourier normalization changes only the nonzero constant factor, not the vanishing conditions.

Use the Fourier transform convention $\hat{h}(\xi )=\int h(x)e^{-2\pi ix\cdot \xi }dx$. The deleted-ball kernel need not be integrable at infinity, so first work with finite annuli. WriteWe will prove a uniform bound, convergence as $R\to \infty$, and then convergence as $\epsilon \downarrow 0$. This makes all Fourier-transform assertions precise for the annularly cancelling singular convolution kernel.

Fix $\xi \ne 0$ and put $y=\xi /(2|\xi {|}^2)$, so $e^{-2\pi iy\cdot \xi }=-1$. For $R>r\ge 4|y|$, let $\Omega =\{r<|x|<R\}$ and defineThe Hörmander integral kernel condition givesOn changing variables $u=x-y$, the phase factor shows $J_{r,R}=-{\int }_{\Omega -y}K(u)e^{-2\pi iu\cdot \xi }du$. The symmetric difference between $\Omega$ and $\Omega -y$ lies in the two shells whose radii are within $|y|$ of $r$ and $R$. On each such shell, the kernel's size bound gives, by polar integration,Thereforeand adding the two comparisons proves the oscillatory tail estimate for a singular kernel:The integral on the right is at most $B$. Moreover, for fixed $\xi$, it tends to zero as $r\to \infty$ because the translated difference is absolutely integrable outside $2|y|$. The shell term also tends to zero. Thus the far Fourier integral is Cauchy and converges as its outer radius tends to infinity.

Set $r_0=4|y|=2/|\xi |$. If $\epsilon <r_0$, annular cancellation allows subtraction of the constant phase on the near annulus:The same estimate applies to a shorter near annulus when $R<r_0$. The far annulus is bounded by $B/2+C_{d}A$, using the tail estimate; if $\epsilon \ge r_0$, apply that estimate with $r=\epsilon$. At $\xi =0$ the annular integral vanishes by cancellation. We have provedFor each fixed $\epsilon >0$, the size bound makes $K_{\epsilon }\in L^2$, and $K{1}_{\{\epsilon <|x|<R\}}\to K_{\epsilon }$ in $L^2$ as $R\to \infty$. By the Plancherel theorem, their Fourier transforms converge in $L^2$ to ${\hat{K}}_{\epsilon }$. The pointwise improper limits just established identify this transform, up to a null set, withFor example identification follows by selecting an almost-everywhere convergent subsequence from the $L^2$ convergence. HenceThe near-annulus integral with the constant phase subtracted converges absolutely at zero, since its bound is proportional to $|x{|}^{1-d}$. The far integral has already converged. Therefore $m_{\epsilon }(\xi )\to m(\xi )$ for every $\xi \ne 0$, with the same uniform bound; the value at zero may be set arbitrarily.

For $f\in L^2$, the convolution $K_{\epsilon }\ast f$ is absolutely convergent for each $x$ by Cauchy-Schwarz inequality. Its Fourier transform is $m_{\epsilon }\hat{f}$: first prove this for Schwartz functions and then approximate any $L^2$ input. The convolutions converge uniformly under this approximation by the same Cauchy-Schwarz bound, and the multiplier outputs converge in $L^2$ by their uniform operator bound, so these two definitions agree. Define the bounded Fourier multiplier operator $T$ by $\widehat{Tf}=m\hat{f}$. Dominated convergence theorem and the Plancherel theorem now give

For the final request, use a normalized bump function, $\int \varphi =1$, so that its dilates form an approximate identity. Multiplying Fourier transforms givesThese products are integrable by Cauchy-Schwarz inequality, since $T{\varphi }_{\epsilon },f\in L^2$, and also define $L^2$ functions because $m$ and $\hat{\varphi }$ are bounded. Thus the convolution identity is justified even though $T{\varphi }_{\epsilon }$ need not be in $L^1$:The allowed standard approximate identity results, applied to $Tf\in L^2$, prove convergence to $Tf$ in norm and almost everywhere. This is the mass normalization in mollified singular integrals; it does not require proving almost-everywhere convergence of the sharp truncations $T_{\epsilon }f$.

If “bump function” means merely smooth with compact support, the stated limit needs the unit-integral qualification. For arbitrary such $\varphi$, the limit is $(\int \varphi )Tf$ instead. This is also the almost-everywhere limit: at each Lebesgue point of $Tf$, subtract $(\int \varphi )Tf(x)$ inside the convolution and bound the remaining local average by the boundedness and compact support of $\varphi$. As a concrete counterexample without normalization, take $K(x)=1/(\pi x)$ in dimension one and $\varphi =2\rho$ for a unit-integral mollifier $\rho$. This kernel has the required size, odd annular cancellation, and translation-difference integral $\log (3)/\pi$ on $\{|x|>2|y|\}$ for every $y\ne 0$. Its limit operator is the nonzero Hilbert transform; by the Hilbert-transform Fourier multiplier, it is an $L^2$ isometry. For any nonzero $f\in L^2$ the mollified limit is $2Tf$, not $Tf$.

For odd $p$, write $e_p(t)=e^{2\pi it/p}$ and use the discrete paraboloidHere surface integration means $p^{-2}{\sum }_P$, whereas ambient norms below use unnormalized counting measure on ${\mathbb{F}}_p^3$. With the negative-phase transform convention,When $x_3=0$, additive-orthogonality of characters makes this one at $x=0$ and zero at all other points of that plane. When $x_3\ne 0$, completing the two squares givesThe Quadratic Gauss sum identity is $G(t{)}^2=\chi (-1)p$, where $\chi$ is the Legendre symbol. For completeness, counting the preimages of a square gives $G(t)=\chi (t)G(1)$ for nonzero $t$. Also $|G(1){|}^2={\sum }_{u,v}{e}_p(u^2-v^2)=p$, by the invertible change $(u,v)\mapsto (u-v,u+v)$ and orthogonality of characters. Since $\overline{G(1)}=G(-1)=\chi (-1)G(1)$, the asserted square identity follows. Thus for $p\equiv 3(\mathrm{mod}4)$,Using positive phase instead changes the sign inside the character, not the magnitude or any norm estimate.

Define $Eg(x)=p^{-2}{\sum }_{\xi \in P}g(\xi )e_p(x\cdot \xi )$, and define $R^{\ast }(2\to q)$ as the smallest constant forThe adjoint relative to these measures is $E^{\ast }h(\xi )={\sum }_{x}h(x)e_p(-x\cdot \xi )$. Consequently $E{E}^{\ast }h=h\ast \hat{\sigma }(-\cdot )$, where convolution is the unnormalized ambient sum. Put $K=\hat{\sigma }(-\cdot )-1_{\{0\}}$. Its supremum is at most $p^{-1}$, so convolution by $K$ has ${\ell }^1\to {\ell }^{\infty }$ norm at most $p^{-1}$.

Under the unnormalized ambient Fourier transform on a finite group, the convolution multiplier of $\hat{\sigma }(-\cdot )$ is $p^3\sigma =p{1}_P$, up to the harmless reflection convention. Thus the multiplier of $K$ is $p{1}_P-1$, with maximum absolute value at most $p$. The Plancherel theorem gives ${\ell }^2\to {\ell }^2$ norm at most $p$. The Riesz-Thorin interpolation theorem between these two bounds gives ${\ell }^{4/3}\to {\ell }^4$ norm at most one for convolution by $K$. Convolution by $1_{\{0\}}$ is the identity, whose ${\ell }^{4/3}\to {\ell }^4$ norm is at most one for counting measure. Hence $\Vert E{E}^{\ast }h{\Vert }_4\le 2\Vert h{\Vert }_{4/3}$.

By Lp duality and Hölder's inequality, $\Vert E^{\ast }h{\Vert }_{L^2(\sigma )}^2=⟨E{E}^{\ast }h,h⟩\le 2\Vert h{\Vert }_{4/3}^2$. Taking adjoints proves the finite-field paraboloid fourth-moment extension estimateIn fact this argument works for both residue classes of odd primes.

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The complete line $\ell =\{(t,it,0):t\in {\mathbb{F}}_p\}$ lies in $P$. For $g=1_{\ell }$ the normalized surface $L^2$ norm is $p^{-1/2}$, and orthogonality of characters gives $Eg(x)=p^{-1}{1}_{\{x_1+i{x}_2=0\}}$. That annihilator plane has $p^2$ points, so the isotropic-line obstruction to finite-field restriction givesThis tends to infinity along these primes for every $q<4$, as required.

We derive the required Calderón–Zygmund decomposition and apply the preceding two parts. Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, and put $A=\Vert \hat{K}{\Vert }_{\infty }$. The Plancherel theorem givesFor completeness this starts with smooth rapidly decreasing inputs, where $\widehat{K\ast f}=\hat{K}\hat{f}$, and extends by $L^2$ density. It agrees with the pointwise convolution in the question: if $f_j\to f$ in $L^2$, then $\Vert K\ast (f_j-f){\Vert }_{\infty }\le \Vert K{\Vert }_2\Vert f_j-f{\Vert }_2$. Thus there is no need to assume that $K$ itself is integrable.

Fix $f\in L^1({\mathbb{R}}^d)$, let $F=\Vert f{\Vert }_1$, and choose $\lambda >0$. Select the maximal dyadic cubes $Q$ for which $|Q{|}^{-1}{\int }_Q|f|>\lambda$. They exist above every qualifying cube because averages over increasingly large ancestors tend to zero. They are disjoint, their total volume is at most $F/\lambda$, and maximality givesOutside their union, dyadic differentiation gives $|f|\le \lambda$ almost everywhere. One can obtain this differentiation fact from the allowed martingale convergence theorem: on each fixed dyadic cube the averages are conditional expectations of the integrable restriction, their almost-everywhere limit exists, and approximation by dyadic step functions shows in $L^1$ that this limit is $f$.

Let $f_Q=|Q{|}^{-1}{\int }_{Q}f$, define $g=f$ outside the selected cubes and $g=f_Q$ on $Q$, and set $b_Q=(f-f_Q)1_Q$. This constructswith the estimatesThese follow by integrating separately on each selected cube; $|f_Q||Q|\le {\int }_Q|f|$ gives both the good-part norm bound and the last cancellation-error bound.

Let $\Omega ={\bigcup }_Q\hat{Q}$, with the dilation $L=4\sqrt{d}$ from part (ii). Then $|\Omega |\le L^{d}F/\lambda$. On its complement, part (i), followed by part (ii), givesFinally, $|Tf|>\lambda$ implies $|Tg|>\lambda /2$ or $|Tb|>\lambda /2$. The first alternative is controlled by the $L^2$ operator bound, and the second by the last integral outside $\Omega$. HenceThus $T$ has weak type $(1,1)$, with a constant depending only on the dimension, the bounded Fourier transform, and the Hörmander integral kernel condition.

Use $\hat{f}(\lambda )=\int f(t)e^{-i\lambda t}dt$ and interpret the regularity assumption to include $f\in L^2(\mathbb{R})$ with the continuous representative furnished by Fourier inversion. Write $F=\hat{f}$. The bandwidth assumption implies $F\in L^1$ as well as $L^2$, by the Cauchy-Schwarz inequality on its finite support, sois continuous and defined at every $t$. The integer samples are precisely the Fourier coefficients of $F$ with the sign convention appropriate to the basis $e^{-in\lambda }$:The completeness and Parseval identity for the exponential orthonormal basis on $[-\pi ,\pi ]$ give $F={\sum }_{n\in \mathbb{Z}}f(n)e^{-in\lambda }$ in $L^2$ andThe last equality is the Plancherel theorem. Apply the inverse integral to the finite symmetric Fourier partial sums. Sincewe obtain the Nyquist–Shannon sampling theorem in this normalization:where the quotient is one at $t=n$. This is a sampling expansion by periodic Fourier projection, not merely an almost-everywhere inversion. Indeed, the inverse integral of an error $G\in L^2[-\pi ,\pi ]$ is bounded, uniformly in $t$, by $\Vert G{\Vert }_2/\sqrt{2\pi }$. Hence the symmetric finite sampling sums converge uniformly on all of $\mathbb{R}$. Moreover, the Parseval identity applied to $e^{it\lambda }$ givesThe Cauchy-Schwarz inequality therefore makes the sampling series absolutely convergent at every $t$, with its absolute tail bounded uniformly by $({\sum }_{|n|>N}|f(n){|}^2{)}^{1/2}$. Thus the integer samples recover the function everywhere.

For the nonuniqueness beyond this bandwidth, fix $0<{ϵ}_0<\min (ϵ,\pi )$ and a nonzero smooth function $\Phi$ compactly supported in $(-{ϵ}_0,{ϵ}_0)$. Let $h$ be its inverse Fourier transform and putThe function $h$, and hence $g$, is a Schwartz function. The two shifted spectral supports are disjoint, so $\hat{g}\ne 0$, but its support is contained strictly inside $(-\pi -ϵ,\pi +ϵ)$. Every integer sample of $g$ is zero. Therefore $g$ and the zero function have identical samples but are distinct, providing a strictly supercritical sampling alias counterexample for every positive bandwidth enlargement.

Use the angular-frequency Fourier transform conventionA precise sufficient interpretation of the regularity assumption is $f\ne 0$, $f\in H^1(\mathbb{R})$ in the first-order Sobolev space and $xf\in L^2(\mathbb{R})$; in particular, nonzero Schwartz functions are admissible. We use the Plancherel theorem, in the normalization $\Vert \hat{f}{\Vert }_2^2=2\pi \Vert f{\Vert }_2^2$, together with its derivative identity $\widehat{f^{\prime }}(\lambda )=i\lambda \hat{f}(\lambda )$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for Schwartz functions. Thus the frequency factor isFor a Schwartz function, integration by parts gives $\Vert f{\Vert }_2^2=-2\mathrm{Re}\int x{f}^{\prime }(x)\overline{f(x)}dx$. This remains true under the stated assumptions: insert a smooth cutoff ${\chi }_R(x)=\chi (x/R)$ into the derivative of $x|f{|}^2$, where $\chi$ is one near zero and compactly supported. Integration givesThe cutoff-derivative term tends to zero, since $x{\chi }_R^{\prime }$ is bounded independently of $R$ and supported in a tail. The other terms converge by dominated convergence and the integrability of $|f^{\prime }||xf|$, furnished by the Cauchy-Schwarz inequality. HenceAfter squaring and using the Plancherel theorem, we obtain the uncentred Fourier uncertainty principle:If equality holds, both displayed inequalities must be equalities. The equality condition in the Cauchy-Schwarz inequality gives $f^{\prime }=cxf$ almost everywhere for a constant $c\in \mathbb{C}$; the real-part inequality and the identity for $\Vert f{\Vert }_2^2$ then force $c=-a$ with $a>0$ real. The weak derivative equation has the solution $f(x)=C{e}^{-a{x}^2/2}$: locally multiply by $e^{a{x}^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this Gaussian function the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. ThereforeThese are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the Gaussian function are not equality cases here; the equality case of the Heisenberg uncertainty relation for centred variances has those additional freedoms.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.