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Another hint for a counterexample to (1): Think of $u$ as point-symmetric function, i.e. $u(x)=-u(-x)$. Let $\phi$ be a non-negative testfunction with $\int\phi=1$ and $\phi(x)=\phi(-x)$. Then $u(\phi)=0$ but it is not difficult to find a test function $\psi$ with $u(\phi\psi)\neq 0$.

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Hint: Start writing down what $\phi u=0$ means. It means that $\phi u$ is the zero distribution, i.e. for all test functions $\psi$ it holds that $(\phi u)(\psi) = 0$ and by definition of the product of smooth functions with distributions that means that for all test functions $\psi$ it holds that $u(\phi\psi) = 0$.

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The discrete-time Fourier transform (DTFT) is defined as $X(\theta)=\sum_n f[n]e^{-i\theta n}$ and it is the equivalent Fourier transform for discrete time series. The resulting $X(\theta)$ is in continuous time and is $2\pi$-periodic. In your case, $f[n]=1$ and you're asking for $X(\theta)|_{\theta=-\omega T_0}$. The DTFT of $1$ should be a delta ...

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The function $f(\omega) = \frac{T_0}{2\pi}\sum_{k=-\infty}^\infty e^{i k \omega T_0}$ is a Fourier series of $\sum_{k=-\infty}^\infty \delta\left(\omega - \frac{2\pi k}{T_0}\right)$. This function is most commonly known under the names the Dirac Comb and the Shah Function.

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Joelafrite made a good suggestion: consider the first distributional derivative of a function $f$ that is not absolutely continuous. By definition, this distribution acts as $$\phi\to -\int f\phi'$$ If $f$ is increasing (like Cantor staircase and Minkowski's ?-function), then the distribution $f'$ is a measure. If $f$ has bounded variation, then $f'$ is ...

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You say you get to $$-\int_{-l}^l e^{ikx} \phi''(x) \, \mathrm{d} x$$ but then apply the Riemann-Lebesgue lemma.

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The sentence "The Kullback–Leibler divergence is defined only if $Q(i)=0$ implies $P(i)=0$ for all $i$" implies that $D_{KL}(P\|Q)$ is not defined if there is some $i$ such that $Q(i)=0$ but $P(i)\not=0.$ One could try to finagle a definition for $D_{KL}(P\|Q)$ in these cases using limits, as is done when $P(i)=0$ for some $P(i)$. The relevant limit (using ...

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How about using the duality? For T smooth enough, you have $$\langle \psi(D)T, \varphi \rangle = \int_{\mathbb{R}} [\psi(D)T](x) \varphi(x) dx$$ $$= \int_{\mathbb{R}} \varphi(x) \int_{\mathbb{R}} e^{i x \xi} \psi(\xi) \hat{T}(\xi) \, d\xi dx$$ $$= \int_{\mathbb{R}} \frac{1}{2 \pi} \psi(\xi) \hat{T}(\xi) \int_{\mathbb{R}} e^{i x \xi}\varphi(x) \, dx ... 1 This is similar to the fact that pointwise convergence does not imply L^1 convergence. Construct a sequence f_n, each with integral 1, that converges to 0 pointwise. 1 f_n \to f in D' if$$\forall \phi \in D, \ \lim_n \int_U (f-f_n)\phi = 0$$If the convergence is strong in L^p, Hölder inequality gives us the result :$$\left|\int_U (f-f_n)\phi \right| \leq \int_U \left|(f-f_n)\phi \right| \leq \|f_n-f\|_p\|\phi\|_q \to 0$$If the convergence is weak, as D\subset L_q, the answer is immediate 0 Take a test function \phi\in D(U) and consider the f_n as distributions:$$|\langle f_n,\phi\rangle-\langle f ,\phi\rangle|=\left|\int_U (f_n(x)-f(x)\phi(x)dx\right|\le \int_U |f_n(x)-f(x)||\phi(x)|dx \le \|f_n-f,L^p(U)\|\|\phi,L^q(U)\|$$with \frac 1p+\frac 1q=1 by Hölder's inequality. Therefore, if f_n\to f in L^p, then the above difference ... 2 Take any compact K not containing zero, then take any test function \phi with support in K, then$$\langle pv(f),\phi\rangle = \lim_{\epsilon\to 0}\int_{\|x\|\ge \epsilon}f(x)\phi(x)dx.$$If$$\epsilon< dist(K,0),$$then$$ \int_{\|x\|\ge \epsilon}f(x)\phi(x)dx = \int_Kf(x)\phi(x)dx,$$hence$$\langle pv(f),\phi\rangle = \int_{K}f(x)\phi(x)dx$$and ... 0 In general, if \Omega\subset\mathbb{R}^n is simply connected and \vec F=(F_1,\dots,f_n)\colon\Omega\to\mathbb{R}^n is a vector field of class C^1, then \vec F is a gradient if and only if$$ \frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i},\quad1\le i,j\le n.\tag{*} $$If the F_i are polynomials, then they are C^\infty on ... 1 As you did the first part, I do not address it. Concerning the examples, c(\sin z)/z is the first example that you need. For the second example, take \phi(t) infinitely differentiable, with support on (-1,1), and consider the function$$f(z)=\int_{-\infty}^\infty\phi(t)e^{izt}dt. This is evidently bounded: $|f(x)|\leq \|\phi\|_1$. Now, ...

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