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Consider an ideal coherent monochromatic wave from an unresolved source on the optical axis, a uniform circular entrance pupil of diameter $D$, and ideal relay optics without wavefront errors. In the scalar Fraunhofer diffraction approximation, a Fourier transform carries the pupil's complex amplitude into the image plane. Choose dimensionless pupil coordinates $\mathbf{x}$ in units of $D$ and angular coordinates $\mathbf{u}=D\mathbf{\theta }/\lambda$, where $\lambda$ is the wavelength. DefineThe coronagraphic focal plane mask has transmission $M=1-W$, with $W=1$ inside the occulting disk and $W=0$ outside. Thus the field immediately after the mask is $MA$. Outside the disk $M=1$, so the local wave intensity is still $|A{|}^2$: hiding the core has left those rings intact. Reimaging an unlimited pupil giveswhere $\ast$ is convolution, $L$ is the Lyot stop transmission, and the proportionality constant in the irradiance is omitted. Coordinate reversals from physical relay optics are absorbed into the coordinate choice. These equations use wave amplitudes throughout propagation; applying a Fourier transform to wave intensity would give a different and incorrect model. This is the pupil/image calculation described by Sivaramakrishnan et al.; the circular disk and numerical choices below are our illustrative specialization. ^[ref].

The subtraction explains the beauty of the arrangement: in much of the interior, the original pupil field and the convolution term nearly cancel. Near a boundary, that cancellation is incomplete. The Lyot stop can therefore block a region containing disproportionately much unwanted light. A hard-edged coronagraphic focal plane mask has a transform with extended tails, so the interior need not be perfectly dark. With an unlimited relay, $L=1$ gives $A_{\mathrm{final}}=MA$ exactly: no further ring suppression occurs. A stop covering the full original optical pupil instead uses $L=P$ and already clips field outside the optical pupil; an undersized Lyot stop clips more of the remaining rim.

The toy uses a coronagraphic focal plane mask of angular radius $3\lambda /D$ and a Lyot stop of diameter $0.8D$. The discrete Fourier transform grid has $1024\times 1024$ samples, with 128 across $D$ and angular sampling $0.125\lambda /D$; it approximates the continuous equations on a finite periodic domain. The plot averages wave intensity in annuli of width $0.25\lambda /D$. Over the annulus $4\le |\mathbf{u}|\le 10$, the mean wave intensity divided by the direct stellar peak is approximately $1.40\times 10^{-4}$ without a coronagraph, the same with only the focal mask and an unlimited relay, $4.44\times 10^{-5}$ with the mask and full pupil, and $1.66\times 10^{-6}$ with the mask and $0.8D$ Lyot stop. Thus the undersized opening reduces this particular annular mean about 27-fold relative to the full-pupil comparison. It does not suppress every direction by that factor, and these values do not predict a real instrument's detection limit.

The ideal toy excludes a central obstruction, support vanes, scattering of light, wavefront errors, atmospheric effects, a spread of wavelengths, pointing drift and detector noise. It also contains only the star: no simulated planet or solar corona. For a far off-axis source whose image misses the coronagraphic focal plane mask, the continuous geometric coronagraph throughput of the $0.8D$ opening tends to $0.8^2=0.64$ relative to the full disk. Rejecting stellar light therefore costs some desired light as well. Enlarging the coronagraphic focal plane mask narrows the characteristic scale of its pupil-plane transform by the scaling property of the Fourier transform, but also hides a larger neighborhood of the source.

The scaling property of the Fourier transform extends by duality toConsequently a homogeneous distribution of degree $\sigma$ transforms into one of degree $-n-\sigma$.

Write $J_{n,h}$ for the squared integral norm in question. The scaling property of the Fourier transform gives $\mathcal{F}{K}_h(u)=\mathcal{F}K(hu)$, which vanishes outside $[-1/h,1/h]$. Also $|\mathcal{F}K(v)|\le \Vert K{\Vert }_1$. The condition $\int K=1$ does not imply $\Vert K{\Vert }_1=1$, because $K$ need not be nonnegative.

Part (b), now for observations of $Y$, gives $\mathbb{E}|{\phi }_n^Y(u)-{\phi }^Y(u){|}^2\le 1/n$. Apply the Tonelli theorem to this nonnegative integrand, use the bound on $1/{\phi }_{ϵ}$, and change variables $v=hu$:The characteristic function ${\phi }_{ϵ}$ is continuous and nonzero on the compact interval of integration, so ${\tilde{M}}_h<\infty$ for every $h>0$. The Markov inequality gives, for every $z>0$,Consequently $J_{n,h}^{1/2}=O_{\mathbb{P}}({\tilde{M}}_h/\sqrt{nh})$, including for any deterministic bandwidth sequence $h=h_n>0$. The stochastic order follows directly from this uniform probability bound. Independence of $X$ and $ϵ$ gives ${\phi }^Y={\phi }^X{\phi }_{ϵ}$, explaining why division by ${\phi }_{ϵ}$ is the Fourier deconvolution operation.

The scaling property of the Fourier transform gives, first for Schwartz functions and then by duality,If the homogeneous distribution $u$ has degree $\sigma$, thenPutting $s=1/t$ yields ${\delta }_s\hat{u}=s^{-n-\sigma }\hat{u}$. Thus the Fourier transform of a homogeneous distribution has degree $\boxed{-n-\sigma }$.

Choose a nonzero $f\in L^p({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ whose transform belongs to $L^q$, for example a Gaussian by the Fourier transform of a Gaussian. For $\lambda >0$, set $f_{\lambda }(x)=f(\lambda x)$. The scaling property of the Fourier transform givesThe assumed estimate, applied for every $\lambda >0$, becomesSince both norms are nonzero, this can hold as $\lambda$ tends both to zero and to infinity only if the powers agree. Thus the scaling necessity for an Lp to Lq Fourier bound yieldsHence $q$ is uniquely the conjugate exponent of $p$.

For a real nonzero $b$, put $s=bt$ in the Fourier transform. Reversing the integration limits when $b<0$ produces the absolute value:This is the scaling property of the Fourier transform. If $b=0$ and $f(0)$ exists, the transformed function is constant in time; its transform is $2\pi f(0)\delta (\omega )$ in the sense of tempered distributions, rather than an ordinary convergent Fourier integral.

For $\mathrm{Re}\alpha >0$, the even exponential is $h(t)=g(t)+g(-t)$ except for an irrelevant single value at zero. Linearity and the scaling property of the Fourier transform at $b=-1$ giveIn particular, for positive real $\alpha$ the transform is real and even, as expected from the symmetry of $h$.