## $L^2$ functions with compactly supported fourier transforms form a Hilbert space

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?

## 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?.