Given a fixed compact subset of $ \mathbb{R}$ , I want to show that square integrable functions on the real line whose fourier transforms are supported in the given compact set form a Hilbert space in the $ L^2$ norm. It was easy to show that the functions form a subspace. However I am stuck at the closedness of the subspace. Could anyone please help me?
Tag: transforms
Broken Usage of Laplace Transforms
Let $ \mathcal{L}$ be the Laplace transform function, and let $ \mathcal{L}\bigl(f(t)\bigr) = F(s)$ .
One of the properties of the Laplace transform is that $ \mathcal{L}\Bigl((t)^nf(t)\Bigr) = F^{(n)}(s)$ , i.e., the $ n$ th derivative of the transform with respect to $ s$ . I was testing this property out a bit.
Given $ ty’+y=0$ , find $ y(t)$ . The solution is very obvious when you use integrating factors: $ $ (ty)’=0$ $ $ $ y=\frac{C}t$ $
Now, let’s try solving this differential equation using a Laplace transform!
$ $ \mathcal{L}\bigl(ty’+y\bigr)=0 \tag{1}\label{1}$ $
The transform of $ f(t)=y’$ is $ sF(s)f_0$ , so then… $ $ \mathcal{L}\big(ty’\big) = \frac{d\left(sF(s)f_0\right)}{ds}=sF'(s)F(s) \tag{2}$ $
Returning back to $ (1)$ , we now have $ sF'(s)F(s)+F(s)=0$
which in turn simplifies to $ $ F'(s)=0$ $ so $ $ F(s)=C$ $ and $ $ f(t)=C\cdot\delta(t)$ $
Can someone explain to me why what I did in $ (2)$ isn’t necessarily correct, because that’s the step I’m iffy on right now. Thank you!
Solve using Laplace transforms
I want to solve the PDE
$ $ x\frac{\partial u}{\partial t} +\frac{\partial u}{\partial x}=x$ $ $ $ u(x,0)=0, x>0$ $ $ $ u(0,t)=0, t>0$ $ for $ x>0,t>0$ using a Laplace transform in t.
I Laplace transform the equation and get $ $ xs\hat{u}(x,s)+\frac{\partial}{\partial x}\hat{u}(x,s)=\frac{x}{s}$ $ where $ \hat{u}(x,s)=L(u(x,t))=\int_{t=0}^{\infty} e^{st}u(x,t)dt$ .
Solving this ODE, I obtain $ $ \hat{u}(x,s)=\frac{1}{s^2} + A(s)e^{\frac{sx^2}{2}}.$ $ I then apply the second boundary condition to get $ $ \hat{u}(x,s)=\frac{1}{s^2}\frac{1}{s^2}e^{\frac{sx^2}{2}}.$ $ Now I need to inverse transform to get $ u(x,t)$ . I know that $ $ L^{1}(\frac{1}{s^2})=t$ $ but I don’t know how to find $ $ L^{1}(\frac{1}{s^2}e^{\frac{sx^2}{2}}).$ $
Pushing Cuckoo Eggs under Inverse Radon Transforms
Essentially the inverse of the Radon transforms $ Rf(L)=\int_L{f(x)dx}$ has the ability to reconstruct $ f(x)$ from the integrals over all lines; or, expressed differently to (re)construct $ f(x)$ from a functional, that maps lines to real values.
Essentially this question is aimed at cuckoo eggs the inverse Radon transform will hatch if pushed under it:
Question:
what are examples of other functionals, that map lines to real values, to which the inverse Radon transform can be applied beneficially?
I am especially interested in examples that have been mentioned in publications. Two examples of such line functionals that would be possible candidates, are

mapping lines to the geodesic distance between the pair of points, in which they intersect the boundary of a compact, convex subset of $ \mathbb{R}^n$ (lines that miss the boundary or are tangent to it are mapped to $ 0$ )

mapping lines to their moments of inertia w.r.t. the model of a compact physical object; lines missing a sufficiently small compact containing volume will be mapped to $ 0$
Bing Transforms can only be used with paid versions of Maltego
I was learning about how to do social engineering then i was found some tool called maltego i wanted to play around with it but when i make website transform it says
Bing Transforms can only be used with paid versions of Maltego
I wanted to know how can I use them with free version because the course I am following didn’t tell me about this. The course I am following is of Udemy(Zaid Sabih).
Deconvolution using Fourier transforms
I have a 2D signal in the form of a function $ g(x_m,y_m)$ given as
$ $ \begin{aligned} g(x_m,y_m) = \ & \int^{\infty}_{\infty} \mathrm{d}x_o \ \mathrm{d}y_o \ \frac{1}{\varepsilon^2} P(x_o – x_m,y_o y_m) \ & \quad \times \int^{\infty}_{\infty} \mathrm{d}x_i \ \mathrm{d}y_i \ R(x_o – x_i,y_o y_i)L(x_i,y_i) \end{aligned} \tag{1}$ $
The integrals in equation $ (1)$ can be seen as a convolution of $ P,R$ and $ L$ as $ $ g(x_m,y_m)= ((P/\varepsilon^2)*R*L)(x_m,y_m) \tag{2}$ $
I want to find $ L$ when $ g$ ,$ R$ and $ P$ are given.
I tried to use Fourier transforms to find $ L$ as:
$ $ L = \mathcal{F}^{1} \left[ \frac{\mathcal{F}[g]}{\mathcal{F}[R] \cdot \mathcal{F}[P/\varepsilon^2]} \right]$ $
R[x_, y_] := 0.609739 E^(444.116 (0.00244704 + x)^2  444.116 (0.0322566 + y)^2) + 0.0691803 E^(48.9858 (0.0105298 + x)^2  48.9858 (0.0054951 + y)^2) + 0.66442 E^(426.449 (0.00315949 + x)^2  426.449 (0.0248433 + y)^2); g[x_, y_] := 3.06909 E^(18.585 x^2 + x (13.6144  27.7795 y) + (12.2542  18.1432 y) y) + 0.402245 E^(7.37814 x^2 + x (9.61481  5.57202 y) + (6.46554  6.35048 y) y) + 120.468 E^(0.0245919 x^2 + x (0.00325668 + 0.00197362 y) + (0.00103919  0.0281421 y) y) + 0.773818 E^(3.79704 x^2 + (15.0351  16.3606 y) y + x (1.75472 + 2.86666 y)) + 0.0833316 E^(38.7396 x^2 + (10.6949  14.1731 y) y + x (32.7513 + 18.7984 y)); epsilon = 0.048; P[x_, y_] := E^((2/epsilon)^2 (x^2 + y^2)); FTR = FourierTransform[R[x, y], {x, y}, {u, v}, FourierParameters > {0, 2 \[Pi]}]; FTg = FourierTransform[g[x, y], {x, y}, {u, v}, FourierParameters > {0, 2 \[Pi]}]; FTP = FourierTransform[P[x, y], {x, y}, {u, v}, FourierParameters > {0, 2 \[Pi]}]; InverseFourierTransform[FTg/((1/epsilon^2)FTP*FTR),{u, v}, {xi, yi}, FourierParameters > {1,2*Pi}]
However, the InverseFourierTransform
takes a lot of time and does not return any result. How do I find $ L(x_i,y_i)$ ?, am I doing something wrong here?.