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0

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 $u(\phi\psi)\neq 0$.


1

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 ...


0

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.


1

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 ...


1

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


0

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 ...


0

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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