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0

You can find an elementary treatment in Lighthill's book on Generalised Functions. http://www.amazon.com/gp/aw/d/0521091284/ref=mp_s_a_1_1?qid=1432930439&sr=8-1&pi=SL75_QL70&keywords=Generalised+functions


2

There is no such "function". But in mathematics we use a construction in functional analysis. The "functional" $F$ defined by $F(g) = g(0)$ for all continuous $g$ is the one called "Dirac delta". In physics, they "pretend" there is an actual function, and do computations with it. Since it often produces useful results, there is little reason to stop ...


1

But my question is: what's the proof that the distribution can be constructed in the first place? How can we determine that there exists a function that is zero at all points but one, this exception is at infinity, and that the integration of this distribution across its entire domain is one? Strictly only the last claim holds. $\langle f, ...


5

The Dirac delta is the distribution defined by $$ \langle \delta_{x_0},\varphi \rangle = \varphi(x_0) $$ for all $\varphi \in \mathcal{D}(\mathbb{R}^n)$. It is not a function, so your question cannot be answered.


2

No, pick $n=1$, $f(x)=\xi(x)e^{-|x|}$ and $g(x)=e^{x^2}$, where $\xi \equiv 0$ near $x=0$. But even $f(x)=e^{-x^2}$, as Daniel Fischer points out in a comment. Here the point is that any continuous function is locally integrable, and $f$ cannot compete with functions that blow up at infinity faster than polynomials.


2

It helps to recall Which functions are tempered distributions? — exactly the distributional derivatives of continuous functions of polynomial growth. The full statement of this theorem isn't needed here, but it gives an idea where to look for weird tempered distributions that are represented by a locally integrable function: take the derivative of a ...


6

No, this doesn't follow. Fix a smooth bump function $\varphi$ supported in $[-1,1]$ and let $$ f(x) = \sum_{n=1}^\infty \phi(2^n x-n) $$ The $n$th term is supported in $[n-2^{-n}, n+2^{-n}]$; these supports are disjoint. The series converges in every $L^p$ space. The function $f$ does not tend to zero at infinity, through. On the Fourier side, $$ \hat ...


1

Although I think @NikitaEvseev's answer is the most natural, perhaps the small exercise of rewriting integration over an interval $[a,b]$ as a distribution is worthwhile. That is, consider $$ u(\varphi) \;=\; \int_a^b \varphi(x)\;dx $$ Edit: simpler than what I wrote before: $$ u(\varphi) \;=\; \int_a^b \varphi(x)\,dx \;=\; \int_{-\infty}^\infty (H(x-a) - ...


2

Suspect one could not find continuous $g(x)$. I state that $g(x) = x\cdot H(x-\pi)\cdot H(2\pi-x)$. Namely $g(x) = x$ if $x\in[\pi, 2\pi]$ and $g(x)=0$ for $x\in R\setminus [\pi, 2\pi]$. So $$ \int_{\pi}^{2\pi}x\phi(x)dx = \int_{-\infty}^{+\infty}g(x)\phi(x)dx. $$


1

Here's a summary of, and extending remarks and explanations to, the comments. Note that my description of $u(t,x)$ is based on experimental observations, not formal proofs. I plotted the sum for $u(t,x)$ for a few values of $t$: within a moderate range for $x$, it would converge, although I did need extended accuracy (used Maple for this). My understanding ...


0

Hints: You may notice the following facts: $$ \text{erf}(-x)=-\text{erf}(x),\\ \lim_{x \rightarrow +\infty }\text{erf}(x)=1 $$ Therfore: $$ \lim_{x \rightarrow \infty }1-\text{erf}(x)=1-\text{sign}(x) $$ Can you take it from here? Edit: Another way to prove this would be to use the following representation of the $\delta$-distribution ...


1

Both spaces allow for distributional derivatives, which is one key reason for defining distributions. But there is a big difference between the two. The key difference, and why we use Schwartz space is that it allows us to define the Fourier transform on distributions. Since $C_c^{\infty}(\mathbb{R})$ is smaller than $S$, ...


1

Graphically, what this function looks like is essentially a bump. On the set $|x|\leq n$, it looks like a plateau that quickly drops off to 0 as $|x|\to n$. So naturally, as $n\to \infty$, the function is essentially going to approach a constant function, and since the height of the plateau is proportional to $1/n$, this has to approach the $0$ function (see ...


0

This first result may be known to you already, but I include it here for the benefit of other users. Theorem 1. If $f:\mathbb{R}^{d}\rightarrow\mathbb{C}$ be a locally integrable $1$-periodic function, then $f$ defines a tempered distribution by integration. Proof. Let $Q$ denote the unit half-open cube $[0,1)^{d}$, and let $Q_{k}:=Q+k$ for ...


0

You can just differentiate both parts again to get: $$ f''(x) = \begin{cases} -\frac{1}{4x\sqrt{x}}, & \text{if $x>0$} \\ -\frac{2}{9\sqrt[3]{x^2}|x|}, & \text{if $x<0$ } \end{cases} $$


2

There is an issue here which is that $\mathcal{S}\cup\{1\}$ is not a vector space so we can't talk about linear functionals on $\mathcal{S}\cup \{1\}$. If $f\in\mathcal{S}-\{0\}$, then $f+1\notin\mathcal{S}\cup \{1\}$. And even if we change the question by instead taking the vector space generated by $\mathcal{S}$ and $1$ and defining the topology blah ...


1

In this answer I will assume that $h$ is defined on $\mathbb R$ instead of $[0,1]$ and with $\int h <\infty$ (so that I do not need to write $\cap [0,1]$ all the time). Also I will make an extra assumption on $h$ (You will see). For any $d>0$ we have \begin{equation} \begin{split} \bigg|\int_{\mathbb R} &h_\lambda(x) f(x) dx -f(a) \bigg| ...


0

The derivative of the delta "function" is a distribution but not a probability measure; to make $L(1)=1$, add your favorite probability distribution to it. For example, $\delta'(x)+\delta(x)$.


2

A summary of the proof in the link: Fix a test function $\eta$ with integral $1$ and define $C=\langle T,\eta \rangle$. You have to do this somewhere to identify the constant. Then, given a test function $\phi$, find a test function $\psi$ such that $$\psi'(x)=\phi(x)-\eta(x) \int_{-\infty}^\infty \phi(y) dy.$$ One thing the proof in the link does not ...


0

Intuitively you a solution to $x^2 u = \delta_0$ is a distribution of the form : "some distribution that is killed by the multiplication by $x^2$" + "a particular solution to $x^2u = \delta_0"$. That is to say that a general solution is "sln to $x^2u = 0$" + "sln to $x^2u = \delta_0$". Solution to the homogeneous system $x^2u = 0$ What does a solution to ...


1

$$\begin{align} \int_{-\infty}^{\infty}(f_n(x)-f(x))\phi(x)dx&=\int_{0}^{\infty} ((f_n(x)-f(x))\phi(x)+(f_n(-x)-f(-x))\phi(-x))dx\\\\ &=\int_{0}^{\infty} (f_n(x)-f(x))(\phi(x)-\phi(-x))dx \end{align}$$ where we used the fact that $f_n$ and $f$ are odd to arrive at the last equality. Now, $f_n(x)-f(x)=-\frac{e^{-nx}}{\cosh(nx)}$. Let's look at the ...


3

Yes, it does. Note that for any $\delta > 0$, convergence is uniform outside $(-\delta, \delta)$. For any test function $\phi$ and $\epsilon > 0$, take $\delta$ so $\int_{-\delta}^\delta |\phi(x)|\; dx < \epsilon$. Then if $n$ is large enough that $|f_n - f| < \epsilon$ outside $(-\delta,\delta)$, $$\left|\langle T_{f_n}, \phi \rangle - ...


1

You have that $f_n \phi \to f \phi$ pointwise As $|f_n (x) \phi(x)| \leq |\phi(x)|$, $f_n \phi$ is dominated by $|\phi|$, that is an integrable fontion. Hence, all the hypothesis of the Lebesgue's Dominated Convergence Theorem are satisfied, and this imply that $$\lim_{n\to +\infty} \int_{\mathbb{R}} |f_n (x) \phi(x)-f(x)\phi(x)| dx = 0$$


0

In the multiindex $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ not all components are equally important. Since you integrate over the first variable, having $\alpha_1 >0$ makes the distribution identically zero: the integral over $x$ of $x$-partial derivative simplifies by the fundamental theorem of calculus, and turns out to be $0$ because $\varphi$ is ...


1

Hint: $$|\tanh(x)-1|=\left | \frac{e^x-e^{-x}}{e^x+e^{-x}}-1 \right |=\left | \frac{e^x-e^{-x}-e^x-e^{-x}}{e^x+e^{-x}} \right | \leq e^{-x}$$ The situation is exactly the same on the other side, since $\tanh$ is odd.


1

Hint We have an odd function here: $$\tanh -x = -\tanh x$$ This gives rise to $$\int_0^\infty (\tanh nx - 1)\phi(x) - (\tanh nx - 1)\phi(-x)\ \mathrm dx = \int_0^\infty (\tanh nx - 1)(\phi(x) - \phi(-x))\ \mathrm dx$$ The second factor is another testing function, so all you need to do is show that $$\int_0^\infty \tanh nx - 1 \ \mathrm dx \to 0$$



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