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Let $C$ be the part of the plane where $x\lt y$. Then our probability is $$\iint_C f_{X,Y}(x,y)\,dx\,dy,$$ where $f_{X,Y}(x,y)$ is the joint density function of $X$ and $Y$. So the region of integration is the region above the line $y=x$. And yes, we assume that $X$ and $Y$ are independent. Express as an iterated integral. It is convenient to integrate ...


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Your formulas are correct if $f$ is measurable and integrable (or measurable and positive). The delta function is not an actual function. If you try to represent it as a function, it should vanish outside the origin. But the origin has measure zero so it can be neglected as you justified. You have just managed to prove that there is no $L^1$ function with ...


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The two forms "are equivalent" (that is, the two functions are the same) whether $$u_1(x,t)-u_2(x,t)=u_\infty(x)=0.$$ It seems this equality is valid because, taking $b>t$, we get \begin{align} u_1(x,t)&=\int_{t}^\infty\partial_s u(x,s)\ ds\\ &=\int_{t}^b \partial_s u(x,s)\ ds+\int_{b}^\infty\partial_s u(x,s)\ ds\\ &=\int_{t}^b \partial_s ...


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I also posted this question on MathOverFlow and Paul Garrett gave the answer I was looking for here.


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This answer is primarily to expand on this comment. From that comment and the following, it seems to me draks thinks of the delta as a function, and from the title, it seems the OP also does. Or at least, this was true at the time of posting the comment and the question respectively. Eradicate this misconception from your minds at once, if it is there. The ...


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Since $\varphi\in C_0^\infty(0,T)$, we also have $\varphi'\in C_0^\infty(0,T)$. The weak convergences $u_k\rightharpoonup u$ and $v_k\rightharpoonup v$ imply that $$ \lim_{k\to\infty}\int_0^Tu_k(t)\eta(t)dt=\int_0^Tu(t)\eta(t)dt $$ and $$ \lim_{k\to\infty}\int_0^Tv_k(t)\eta(t)dt=\int_0^Tv(t)\eta(t)dt $$ for all $\eta\in C_0^\infty(0,T)$. (If you defined the ...


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If you know your distribution up to a constant, a good way to fix the constant is to pair the distribution against a test function $f$. For simplicity, we can pick such an $f$ that both $f$ and $F(f)$ are real and symmetric (a Gaussian, for example). Now calculate $\langle F(u),{F(f)}\rangle$ in two ways: $$ \langle F(u),F(f)\rangle = p.v.\int\frac1{2\pi ...


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Ok, from a comment above, I assume that this is what you should do in order to have your operator as a Fourier Multiplier one (or, more essentially, as a well-defined one!) STEP 1: First, you should correct a bit the notion of $f * \frac{e^{it \alpha}}{t} $: this, in general, does not converge, so that, to make clear what this means to be, you should ...


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You don't really need distribution theory to see this, but some topology. The weak*-topology of $C[-1,1]$ is the topology induced by the maps $\Gamma_f: C^*[-1,1] \to \Bbb{R}$, $\Gamma_f(\Lambda) = \Lambda(f)$. Your book is claiming that this topology is the topology of pointwise convergence, meaning that $\Gamma_i \to \Gamma$ in $C^*$ if and only if ...


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I'd say his definition doesn't define what $\delta$ is, only what "converging to $\delta$" means. Just like in analysis, you say the sequence $a_n$ is said to converge to $+\infty$ if $$\forall M \in \mathbb{R},\ \exists N \in \mathbb{N},\ \forall n > N,\ a_n > M$$ This doesn't define what $+\infty$ is. Now he can prove that this definition of ...


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Since $\delta$ is a distribution, the shorter answer is, yes you do need distribution theory to understand this. However, if you are only concerned with the identity if $\int_{-1}^1 f(x)k_n(x) dx = f(0)$ for $f \in C([-1,1])$, then the right hand side is just $\delta(f)$. On the left hand side, you probably have to use the definition of $k_n$, change of ...


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Generally, what do we mean by $T(\lambda x)$ when $T$ is a distribution? To define $T$, we must specify what it does to test functions. This must be done in a way that is consistent with what happens when $T$ is an ordinary function. In the latter case $$ \int T(\lambda x)\phi (x)\,dx = \int T(y)\phi (\lambda^{-1} x)\,\lambda^{-1} dx = \lambda^{-1} \langle ...


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$Q1.$ According to my school's EE graduate course catalog, there are a couple of classes covering Hilbert, Banach and L^p spaces; and linear functionals. Is it possible to cover these topics without a formal mathematical background? $A1.$ It is the responsibility of the school to provide a curriculum for the students that is both meaningful an doable. ...


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Your solution is correct, apart from the fact that you forgot the scale factors for the Dirac terms. It should be $$\hat{T}(k)=\frac{1}{\sqrt{2\pi}}\left( \frac{2i\beta -\lambda k-\mu ik^2}{k^3}\right)+\sqrt{\frac{\pi}{2}}\left( \mu\delta(k)+i\lambda\delta'(k)-\beta\delta''(k)\right)$$ It looks like Mathematica can't be trusted here. First of all, it gets ...


2

Well, derivation under the integral sign gives you directly $$\delta(y)=\int e^{-ixy}dx$$ $$\delta'(y)=-i\int xe^{-ixy}dx$$ $$\delta''(y)=-\int x^2 e^{-ixy}dx$$ $$\delta'''(y)=i\int x^3 e^{-ixy}dx$$ Then, by integration by parts, you can verify what it does as a distribution: $$\int \delta(y)f(y)dy=f(0)$$ $$\int ...


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There are many equivalent ways to construct the topology on $C^\infty_c(\Bbb{R^d})$ that makes it an LF-space. I'm not familiar with Tao's method, but I've looked at a different construction that should be equivalent, and is in my opinion very intuitive. First, we set our goal to be to find a topology on $C^\infty_c(\Bbb{R^d})$ that makes it a locally ...


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I'm new around here (so suggestions about posting are welcome!) and want to give my contribution to this question, even though a bit old. I feel I need to because using the divergence theorem in this context is not quite rigorous. Strictly speaking $1/r$ is not even differentiable at the origin. So here's a proof using limits of distributions. Let ...



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