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2

The weak derivative is $(y-x) \delta(t)$, assuming $x,y$ are fixed.


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The statement $$\delta(x) = \lim_{v\to 0}\frac{e^{-x^2/2v}}{\sqrt{2\pi v}}$$ means that for every $C^\infty$ smooth function $\varphi$ with compact support we have $$\varphi(0) = \lim_{v\to 0} \int_{-\infty}^\infty \frac{e^{-x^2/2v}}{\sqrt{2\pi v}} \varphi(x)\,dx \tag{1}$$ (I would not expect computer algebra systems to correctly handle various modes of ...


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Try to write $k \mapsto (z- 4 \pi^2 |k|^2)^{-1}$ as the Fourier transform of a function $h\in L^1$ (try a multiple of $x \mapsto e^{-\alpha |x|}$, with $\alpha$ a square root of $-z$ - notice the condition on $z$ is what ensures you can pick a square root with strictly positive real part which is needed to get it to be in $L^1$), then recall $$\mathcal F h ...


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Yes, $\psi$ is $C^\infty$-smooth because Quotient of two smooth functions is smooth. If $\varphi(0)=0$, then $\psi$ has compact support, because in this case $\psi(x)=0$ whenever $\varphi(x)=0$. But if $\varphi(0)\ne 0$, then $\psi $ does not have compact support. Indeed, for large enough $x$ we have $\varphi(x)=0$ and therefore $\psi(x)=-\varphi(0)/x$.


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In fact, no, the method of proof is slightly different. You don't need to find a sequence of test functions that converges to zero in $C^\infty_c(\Bbb R)$. Suppose the contrary: there exists a distribution $u\in D'(\Bbb R)$ such that it's restriction on $\Bbb R^\ast$ is represented by $\exp(1/x^2)$. Then, by definition of distribution, we can fix a compact ...


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Denote by $B_m$ the open ball in $R^d$ of radius $m$, and let $\bar B_m$ denote the closed ball of radius $m$. Given a countable collection of non-empty open neighborhoods of $f=0$, say $U_n$, fix, for each $n$, a non-zero function $f_n\in U_n$ such that $f_n$ has support in $\bar B_{m_{n+1}}\setminus B_{m_n}$, where $m_n$ is a strictly increasing sequence, ...


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Hint: if a function and all its derivatives have at most polynomial growth, then it defines a tempered distribution. Now check that $g(x)$ satisfies the above hypothesis.


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To answer this properly, one has to talk about the reason for introducing distributions first. One of the main reasons for this is to be able to define a sort of generalized derivative for objects which are to "singular" to possess derivatives in a classical sense. For example, the functions $$ f\left(x\right)=\begin{cases} 0, & x<0,\\ 1, & ...



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