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$G$ itself is a function, but the derivative $\Delta_y$ is taken in the sense of distributions, so the resulting object $\Delta_y G(x,y)$ need not be a function. (A simpler example: the Heaviside function $H(x)$ is a function, and its derivative in the ordinary sense is zero for all $x\neq 0$ and undefined at the origin. But one can instead interpret the ...


2

First of all note that $\ln |x|$ is integrable near $0$ so that for any $\phi \in \mathcal{D}'$ the integral $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx$$ is defined, and that $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx.$$ Now, $$\int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx = ...


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I hope to have found the answer. Let $A \subseteq \mathbb{R}^n$ be an arbitrary set and choose an arbitrary open set $\Omega \subseteq \mathbb{R}^n$ with $A \subseteq \Omega$ (e.g. $\Omega = \mathbb{R}^n$). Set $\mathcal{D} := \mathcal{D}(\Omega)$ and $\mathcal{D}' := \mathcal{D}'(\Omega)$. We want to find a pre-dual for the space of distributions $T \in ...


1

Suppose the spikes in the smooth approximation to $\delta'(x)$ are located at $x=-h$ and $x=h$. When $\bar{x} \approx x+h$, the smooth approximation to $\delta'(x-\bar{x})$ will be large and positive, so the integral will roughly pick up "something large" times $f(x+h)$. Similary, for $\bar{x} \approx x-h$, the integral will pick up the same large factor ...


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For $K \subset \mathbb{R}^n$ compact, consider the space $$\mathcal{D}_K := \{ \varphi \in \mathcal{D} : \operatorname{supp} \varphi \subset K\}$$ endowed with the seminorms $$\lVert\varphi\rVert_k = \sup \{ \lvert D^k\varphi(x)\rvert : x \in \mathbb{R}^n\}$$ for $k \in \mathbb{N}^n$. It is straightforward to show that $\mathcal{D}_K$ is then a Fréchet ...


4

Suppose that we have a function $\psi(x)$ so that $$ \delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1} $$ Let $0\le\eta(x)\le1$, $\eta(x)=0$ for $|x|\le\frac12$, and $\eta(x)=1$ for $|x|\ge1$. Define $\phi_\lambda(x)=\psi(x)\eta(\lambda x)$. Then, since $\phi_\lambda$ is $0$ in a neighborhood of $0$, we have $$ \begin{align} 0 ...


1

You have the equation $$e^x(e^{-x}u)'=\delta_0+1.$$ First step is to notice that the function $x\to e^{x}$ is $C^\infty$, strictly positive everywhere, hence we can safely divide by it both sides of the equation without producing and/or losing solutions. Thus, we get $$ (e^{-x}u)'=e^{-x}\delta_0+e^{-x}=\delta_0+e^{-x}.$$ It is easy to take the ...


0

(I would usually post this as a comment, but apparently I need an account for that.) Suppose that $f \in \mathscr{S}'(\mathbb{R}) $ is a solution to the equation $$ f'-f = \delta_0 + 1. $$ Since you are following Gerd Grubb's textbook, have a look at the technique described on page 108, starting around equation (5.41). We can set the differential operator ...


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ok, I think I've figured it out: in my course, while showing that distributional solutions of $$\Delta u = 0$$ are smooth we've shown that you can estimate (locally at least) $|| u||_{H^{k+1}} $ norm by the $|| u ||_{H^k}$ norm, and that works for any $k$ - negative as well. The proof can be also easily adapted to the case $$ \Delta u = f $$ with $f \in ...


1

Not a complete answer, but it increases our knowledge nevertheless. If the sequence converges in $\mathcal S$, then it converges pointwise. Therefore, if $a=0$, we have the following cases: $b=0$ nothing interesting here. The sequence is constant. $b<0$ the sequence converges pointwise to a constant function $f(x)=1\quad \forall x$ $b>0$ the ...


0

How about applying the following identity $$\sum_{n\in \mathbb{Z}} f(n) = f(0) + \sum_{n=1}^\infty [f(n) + f(-n)]$$ to both sides of the Poisson formula $$\sum_{n\in\mathbb{Z}} F(2\pi n) = \frac{1}{2\pi} \sum_{k\in\mathbb{Z}} \hat F(k).$$ We obtain immediately $$ F(0) + \sum_{n=1}^\infty [F(2\pi n) + F(-2\pi n)] = \frac{1}{2\pi} \hat F(0) + \frac{1}{2\pi} ...


1

For $n\geq 3$ take $f_k(x)=\min\{ |x|^{-1}, k\}$. Then it's easy to see that $\| f_k\|_{H^1_0(\Omega)}$ is uniformly bounded in $k$, but $f_k(0)=k$ is unbounded. For $n=2$ a similar argument applies to $f_k(x)=\min\{ \ln \ln(1+1/|x|), k\}$.


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I would say you actually have the boundary terms in $R^N$ case, but it is just disappears because your data function $f$ is compact supported. Moreover, I suggest you work out the convolution solution for the case that $f$ is only integrable and Holder continuos, then you will have a better understanding how the "boundary" of $R^N$ disappears.


5

The sequence $\{A_n\}$ need not to be bounded. To see this, one could for example as $f(t,T)$ choose something that approximates a derivative of a delta distribution as $T\to+\infty$. I wish to give credits to my colleague Tomas Persson who came up with that idea. I will give such an approximating example. My example is non-smooth, but that is just to make ...


1

$0 \lt \mathrm e^{-nt} \le 1 \quad \forall x \in [0,1] \\ \implies A_n \lt \lim_{T \to \infty} \int_0^1 f(t,T) \,\mathrm d t$ so it's bounded above if the limit exists. A similar result gives a lower bound if the limit is negative.


1

Your counterexample does not work: $$\lim_{x\to 1,\,x<1}\psi(x) = \lim_{x\to 1,\,x<1} e^{1 - \frac 1{x^2 - 1}}=+\infty,$$ $$\lim_{x\to 1,\,x>1}\psi(x) = \lim_{x\to 1,\,x>1} e^{1 - \frac 1{x^2 - 1}}=0,$$ hence $\psi$ is not a continuous function. The key point is to understand that The product of two $C^\infty$ functions is also a $C^\infty$ ...


2

First step: prove that the distribution with zero derivative are constant functions. Second step: learn to solve the equation $T'=\delta_0$ (easy to solve; if you can't, search math.SE, there's a question about is under the 'distribution-theory' tag). Third step: learn to solve the differential equation $S'= G$, where $G$ is a piece-wise continuous ...


2

Let us denote $S= T'$, then you have $$id\cdot S = 0,$$ which, as you already solved, yields $$S = c\delta_0.$$ Now you have $$S=T'= c\delta_0.$$ Apparently, now we need to solve this differential equation. First of all, you should, by now, be able to guess a particular solution to this problem: if $H$ - the Heavyside function ...


1

Note that $\dfrac{d}{dx} |x|^{-1/2} = - \text{sgn}(x) |x|^{-3/2}/2$ (where sgn is the signum function), not $- |x|^{-3/2}/2$. $T'(\phi)$ is the Cauchy principal value of $- \displaystyle\frac{1}{2} \int_{-\infty}^\infty \text{sgn}(x) |x|^{-3/2} \phi(x)\; dx$


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Hint: Break the integral into two parts - from $0\to\infty$ and $-\infty\to 0$.


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First, and quite important thing: if the function is $L^1_{loc}$, it does not imply that this function is a tempered distribution. Classic counterexample is $e^x\in L^1_{loc}(\Bbb R)\setminus \mathcal S'(\Bbb R)$. In your case the straightforward argument that continuous periodic functions define tempered distributions is more than sufficient. Second, in ...


3

To follow the definitions of Rudin, I'll use the convention that $(\tau_x f)(y) = f(y-x)$. Writing $u'$ for the weak derivative of the distribution $u$, we need to show that for any test function $\varphi \in \mathcal{D}(\mathbb{R})$, $\langle \frac{u-\tau_x u}{x} , \varphi \rangle \to \langle u', \varphi \rangle$ as $x \to 0$. If $y$ denotes the variable in ...


1

This is just due to the regularity of measurable sets, and I think this is more or less the most elementary way of doing it. If you can show that $\displaystyle \int_E f=0$ for all measurable sets you would be done, since we can just take $E$ to be the set where $f$ takes positive or negative values. So we need to show $\displaystyle \int_E ...


2

I see that you have a satisfactory answer, and that's good. Because I suggested you look into mollifiers, I wanted to mention a couple of results that are fundamental to the study of mollifiers. Mollifiers are convolution operators with positive $C^{\infty}_{0}(\mathbb{R}^{n})$ kernels $\varphi$ for which $\|\varphi\|_{L^{1}(\mathbb{R}^{n})}=1$. It turns ...


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The distributional derivative of a piecewise smooth function consists of two terms: pointwise derivative, found as in calculus singular part: if $f$ is discontinuous at $a$, add $(f(a+)-f(a-))\delta_a$; this is done for each discontinuity. You correctly found that $f'$ has no singular part. However, note that the formula for $f'$ should also say that ...


2

This is a special case of the du Bois-Reymond lemma / fundamental lemma of calculus of variations: If $Ω \subset \mathbb R^n$ is open and $f ∈ L^1_{loc}(Ω)$ such that $$ \int_Ω f(x) \phi(x) dx = 0 \quad \forall \phi ∈ C^∞_0(Ω) $$ then $f$ is zero almost everywhere. Now just consider $Ω = \mathbb R$ and take $f$ as the difference of your two functions.



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