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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, $$\hat ... 3 Yes, it does. Note that for any \delta > 0, convergence is uniform outside (-\delta, \delta). For any test function \phi and \epsilon > 0, take \delta so \int_{-\delta}^\delta |\phi(x)|\; dx < \epsilon. Then if n is large enough that |f_n - f| < \epsilon outside (-\delta,\delta),$$\left|\langle T_{f_n}, \phi \rangle - ...

3

You can formalize this: Take two Hilbert spaces $H\subset V$ such that The inclusion is dense (i.e. the image of $H$ in $V$ is dense in the topology of $V$) and, The inclusion is continuous (i.e. $H$ has a stronger norm than $V$). Then, if we identify $V$ with its dual (let's assume the spaces are real, though you can do this in general), we have ...

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 ...

2

You can see the Dirac not only as a distribution, but also as an element of the dual of continuous functions : a Radon measure. This point of view gives sense to $\delta[1]$. And if a measure $\mu$ is absolutely continuous in respect to the Lebesgue measure, $$\mu( \phi ) = \int_{\mathbb{R}} \phi(x) d\mu(x) = \int_{\mathbb{R}} \phi(x) \mu(x) dx$$ (the ...

2

Suppose $d$ is a distribution with compact support $S$ and $f$ is a smooth function, not necessarily with compact support. Define a smooth, compactly supported cutoff function $k$ such that $k=1$ on a neighborhood of $S$. Then $\langle d,kf \rangle$ makes proper sense. One can prove that the value of this expression does not depend on $k$, under the ...

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 $$\int_{\pi}^{2\pi}x\phi(x)dx = \int_{-\infty}^{+\infty}g(x)\phi(x)dx.$$

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

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 $$\int_0^\infty \tanh nx - 1 \ \mathrm dx \to 0$$

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: $$u(\varphi) \;=\; \int_a^b \varphi(x)\,dx \;=\; \int_{-\infty}^\infty (H(x-a) - ... 1 Graphically, what this function looks like is essentially a bump. On the set |x|\leq n, it looks like a plateau that quickly drops off to 0 as |x|\to n. So naturally, as n\to \infty, the function is essentially going to approach a constant function, and since the height of the plateau is proportional to 1/n, this has to approach the 0 function (see ... 1 In this answer I will assume that h is defined on \mathbb R instead of [0,1] and with \int h <\infty (so that I do not need to write \cap [0,1] all the time). Also I will make an extra assumption on h (You will see). For any d>0 we have \begin{split} \bigg|\int_{\mathbb R} &h_\lambda(x) f(x) dx -f(a) \bigg| ... 1 Here's a summary of, and extending remarks and explanations to, the comments. Note that my description of u(t,x) is based on experimental observations, not formal proofs. I plotted the sum for u(t,x) for a few values of t: within a moderate range for x, it would converge, although I did need extended accuracy (used Maple for this). My understanding ... 1 You have that f_n \phi \to f \phi pointwise As |f_n (x) \phi(x)| \leq |\phi(x)|, f_n \phi is dominated by |\phi|, that is an integrable fontion. Hence, all the hypothesis of the Lebesgue's Dominated Convergence Theorem are satisfied, and this imply that$$\lim_{n\to +\infty} \int_{\mathbb{R}} |f_n (x) \phi(x)-f(x)\phi(x)| dx = 0$$1 It helps to recall Which functions are tempered distributions? — exactly the distributional derivatives of continuous functions of polynomial growth. The full statement of this theorem isn't needed here, but it gives an idea where to look for weird tempered distributions that are represented by a locally integrable function: take the derivative of a ... 1$$\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}$$where we used the fact that f_n and f are odd to arrive at the last equality. Now, f_n(x)-f(x)=-\frac{e^{-nx}}{\cosh(nx)}. Let's look at the ... 1 Both spaces allow for distributional derivatives, which is one key reason for defining distributions. But there is a big difference between the two. The key difference, and why we use Schwartz space is that it allows us to define the Fourier transform on distributions. Since C_c^{\infty}(\mathbb{R}) is smaller than S, ... 1 Use integration by parts and the Riesz representation theorem. For the one inclusion, if u\in H^4(I) \cap H^2_0(I), we can integrate by parts twice to obtain$$a(u,v) = \int\limits_I u''(x)v''(x)\,dx = \int\limits_I u^{(4)}(x)v(x)\,dx$$and see that$$\lvert a(u,v)\rvert \leqslant \lVert u^{(4)}\rVert_{L^2(I)}\cdot \lVert v\rVert_{L^2(I)}. For the ...

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