New answers tagged

0

I think the issue is the original pdf: $$ \int_{\mathbb{R}^2}f(x,y)\;dydx=\frac{3}{8}\int_0^2\int_{0}^{2-x}xy\;dydx=\frac{3}{8}\int_0^2\frac{x(2-x)^2}{2}\;dx$$ $$ =\frac{3}{8}\int_0^2\frac{(2-x)x^2}{2}\;dx=\frac{3}{8}\cdot\frac{2}{3}=\frac{1}{4}$$


2

Yes. If $\phi$ is a test function then dominated convergence shows that$$(f*\phi)'=f*\phi'=0,$$so $f*\phi$ is constant. There is a sequence of test functions $\phi_n$ so $f*\phi_n\to f$ almost everywhere.


2

You may use the theory of distributions, and it is true that if the derivative of a distribution is zero, the distribution is constant. On distributions over $\mathbb R$ whose derivatives vanishes


1

Given $f'\in L^1([0,1])$ let us define a function $$g(x) = \int_0^x f'(t) dt.$$ Now for each $\phi\in \mathcal{D}([0,1])$, if we have $$\int_0^1 [f(x)-g(x)] \phi'(x) dx = 0,$$ this implies that $f - g$ is an a.e. constant function (it is a classic result). Since $f-g$ is continuous, when we plug in the point $0$ into $f-g$, we see that the constant ...


0

Let's look at $(f',\varphi)$, where the test function $\varphi$ equals $1$ on $[h,1-h]$, $0\le\varphi\le 1$, and $\varphi$ is supported by $[0,1]$. Since $f'\in L^1$, dominated convergence shows that $(f',\varphi)\to \int_0^1 f'$ as $h\to 0+$. On the other hand, $$ (f',\varphi)=-(f,\varphi') = -\int_0^h f(x)\varphi'(x)\, dx - \int_{1-h}^h f(x)\varphi'(x)\, ...


0

If $\mathcal D (\Omega, \Bbb R^d)$ is the space of vector-valued test functions, there is a topology on it, very similar to the Schwartz topology on $\mathcal D (\Omega)$, that makes it a locally-convex topological linear space. It makes sense, then, to consider its topological dual, $\mathcal D ' (\Omega, \Bbb R^d)$, the elements of which are called vector-...


0

This is standard notation: if $p \in \mathcal D ' (\Omega)$, then $\nabla p \in \mathcal D ' (\Omega, \Bbb R^d)$ is a vector-valued distribution (the terminology is misleading, because it suggests that a vector-valued distribution has vectors as values, which it does not; the term is quite natural, though, in that a vector-valued distribution is a ...


0

For simplicity, assume that $i=d$ and $d \ge 2$. Let $u \in \mathcal D ' (\Bbb R ^{d-1})$. View it as a distribution on $\Bbb R^d$ by tensorizing it with $1$, i.e. $\langle u \otimes 1, \varphi \rangle = \langle u, \int \varphi (\cdot, x_d) \ \Bbb d x_d \rangle$. Then $$\langle \partial _{x_d} (u \otimes 1) , \varphi \rangle = - \left< u, \int \partial _{...


1

A simple counterexample is $f(x)=e^{-\sqrt{|x|}}$ (OK, that's not smooth at $0$, but just smooth it out near $0$ since all we care about is the behavior for $x$ large). Much more generally, suppose $f_0,f_1,f_2,\dots$ are any countable collection of Schwartz class functions. Then we can find a Schwartz class function $f$ such that for all $n$ there exists $...


1

No, this map is not injective. Indeed, the inclusion $C^\infty(\Omega)\to \mathcal{D}'(\Omega)$ mapping a function to the regular distribution it generates is injective and the $i$-th partial derivative of a regular distribution generated by some $f\in C^\infty(\Omega)$ is the regular distribution generated by $\partial_i f$. So if $f,g\in C^\infty(\Omega)$ ...


0

If $f$ is a distribution (1) may not be meaningful pointwise, so presumably what you mean is a weak solution, i.e. $X$ should be a distribution such that for any test function $\phi$, $$ \langle \phi, A X + f - X' \rangle = \langle A^T\phi + \phi', X \rangle + \langle \phi, f \rangle = 0 $$ I think you'll want to assume $A$ is smooth as a function of $...


1

let's take $\Omega = [0,1]$, $\varphi,f \in C^\infty_c((0,1))$. Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$ $\varphi,f$ have their support strictly inside $(0,1)$ so $f'(0)\varphi (0) = f'(1)\varphi (1)= 0$ and $$\langle f'',\varphi \rangle =-\langle f',\...


2

What prevents you from using Kullback-Leibler divergence (KL divergence) as a measure of distance from the uniform distribution? I do agree with you on the fact that KL divergence is not a true measure of "distance" because it does not satisfy (a) symmetry, and (b) triangle inequality. Nonetheless, it can serve as a criterion for measuring how far/close a ...


2

Total variation distance, also known as statistical distance, is a good metric (very stringent). (Note that up to a factor $2$, it's equivalent to $\ell_1$ distance between the vectors of probabilities.) It also has a nice interpretation in terms of closeness of events. $\ell_2$ will be much more forgiving towards small differences, and put the emphasis on ...


-2

Not sure this is the fastest and most elegant solution, but this fact is rather a standard fact from distribution theory. $\textbf{First case}:$ If $\Omega$ is bounded than it is obvious (as you wrote), since set has finite Lebesgue measure. $\textbf{Second case}:$ Set $\Omega$ is open and unbounded. This is not trivial fact (but a good excercise) to show ...


2

Let $f\in C^\infty_c(\Omega)$. Then $f$ is supported in a compact set $K$ and $|f|$ attains a maximum $C$ in this $K$. Thus $$\int_{\Omega} |f|^p dx = \int_K |f|^p dx \le \int_K C^p dx = \text{Vol}(K) C^p.$$ Thus $f\in L^p$ for all $p$. Indeed $C^\infty_c(\Omega)$ is dense in $L^p$ for all $1\le p <\infty$.


0

I'm assuming that $C^\infty_c(\Omega)$ is the space of $C^\infty$ with compact support contained in $\Omega$. If that's the case, then you don't care whether or not $\Omega$ has finite measure, since $$ \int_\Omega |u|^pdx\leq |spt(u)|\cdot\max |u|^p<\infty. $$ where $spt(u)$ is the support of $u$, that is defined as $$ spt(u)=\overline{\{x\in \Omega :...


0

Well yeah it has compact support, hence outside of a compact set $K$ is zero. It's continuos on a bounded, closed set hence its integral must be smaller than $M\cdot m(K)$, where $M$ is the maximum of the function on $K$ and $m$ is the lebesgue measure


1

You can approximate any function in the Lebesgue space arbitrarily well by a differentiable function. Taking a converging sequence of such approximating functions you can show that $ Vf_n $ is Cauchy in the Sobolev space, hence has a limit in the Sobolev space. Since $ V $ is continuous as operator into the Lebesgue space the Sobolev limit equals $ Vf $


0

In distribution theory, many notions are generalizations of what happens in space $L^1_{loc}(\Omega)$. The distributional derivative is an operator of type $D^\alpha : \mathcal{D}'(\Omega) \longrightarrow \mathcal{D}'(\Omega)$, its restriction on $L^1_{loc}(\Omega)$ is what many authors call weak derivative, in this sense, weak and distributional derivative ...


0

One can define $\Delta$ in $H^1$ because $\Delta$ is in divergence form $$\Delta u = \text{div}\nabla u$$ Thus we define $\Delta : H^1_0(\Omega) \to H^1_0(\Omega)^*$ by $$\Delta f (\phi) := -\int_\Omega \nabla f \cdot \nabla \phi.$$ The right hand side is well defined for $f , \phi \in H^1_0(\Omega)$. If $f$ is in $H^2_0$, then integration by part gives ...


3

First the answer for distributions, then tempered distributions: If $f$ is a distribution with $\Delta f=0$ then $f$ is actually a smooth function, which is to say the nullspace of the Laplacian is the space of harmonic functions. Sketch, assuming the domain is $\Bbb R^d$: Fix $\phi\in C^\infty_c$ with $\int\phi=1$, and such that $\phi$ is radial ($\phi(x)$...


2

The trick is explained here at the end of page 1 : If $f$ is continuous and bounded by $C$, with $$A = \sup_x |(1+x^2) \phi(x)|$$ you get $$\sup_x |(1+x^2) \phi(x)f(x)| \le A C$$ i.e. $$|\phi(x)f(x)| \le \frac{AC}{1+x^2}$$ and $\displaystyle F(\phi) = \int_{-\infty}^\infty f(x) \phi(x) dx$ is a tempered distribution since $$\left|\int_{-\infty}^\infty f(x) ...


0

Let us start by proving the inequality in the hint: For fixed $x\in\mathbb{R}^{n}$, we have $$ \partial_y^\alpha(\tau_x\tilde{\psi})(y) = \partial_y^\alpha(\psi(x-y)) = (-1)^{|\alpha|}(\partial^\alpha\psi)(x-y) $$ and $$ (1+|x-z|^2)^N \leq (1+2|x|^2+2|z|^2)^N \leq 2^N(1+|x|^2+|z|^2)^N \leq 2^N (1+|x|^2)^N(1+|z|^2)^N. $$ Hence $$ \sup_{|\alpha|\leq N,y\in\...


1

Since the mapping $$ \iota\colon\mathcal{D}\to H $$ is a continuous linear operator (which is the "generalization" of bounded operators to the setting of locally convex spaces), there is a (continuous) transpose $$ \iota^t \colon H' \to \mathcal{D}' $$ satisfying $$ \langle \varphi, \iota^t(T)\rangle = \langle \iota(\varphi), T\rangle $$ for all $\varphi\in\...


2

Hint $\phi$ has compact support and $e^{\frac{x}{n}} \to 1$ uniformly on compact sets. P.S. Note that your integration by parts approach cannot help: Indeed, since $e^{x/n} \to 1$ on $supp(\phi')$, we have $$\int_{\mathbb R} e^{x/n} \phi'(x) dx \to \int_{\mathbb R} \phi'(x) dx =0$$ with the last equality following from the fact that the antiderivative of $\...


0

As has been mentioned in a comment your question is very broad. One possible answer is to use the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$, it consists of the linear functional continuous with respect to $\mathcal{S}(\mathbb{R}^n)$ which is the Schwartz space. In other words it can be shown that the ...


0

I think your second argument is somewhat going into the wrong direction. You want to show that $H'\subset\mathcal{D}'$, so have to show two things: 1) Given $f\in H'$, you have $f\in\mathcal{D}'$, which follows since $\mathcal{D}$ is continuously embedded into $H$ (that you still have to show). 2) If two functionals $f,g\in H'$ coincide on $\mathcal{D}$, ...


0

I do not have the book of Duistermaat available at the moment, so I cannot compare your proof with the proof in the book. At the moment I see two problems with the proof: 1) Since you have $L(\delta)\subseteq L$ and not $L\subseteq L(\delta)$, I do not see why $(1-\psi)u$ should be a smooth function. 2) One of your distributions $u$ or $v$ might have non-...


0

You are not allowed to do equip $ \mathcal{D}$ with the restriction $ \|\cdot\|_{\mathcal{D}}$ of the norm induced by $⟨⋅,⋅⟩_H$, because if you change the norm on $ \mathcal{D}$ you change $ \mathcal{D}' $. For example condider $ C^{\infty}_{c}$ with the $ L^2 $ norm then the dual is $ L^2$.


2

Let me handle the case $d = 1$ for simplicity of notation. Since $H$ is a Hilbert space, $F$ is represented by an element $g \in H$. Thus, for all $\phi \in \mathcal{D} \subset H$ with $\operatorname{supp} \phi = K$ we have $$ |F(\phi)| = \left| \left<\phi, g \right>_H \right| \leq ||g||_H ||\phi||_H = ||g||_H \left( \int_{\Omega} \phi \cdot \phi + \...



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