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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 $$\mathcal F h ... 1 Yes, \psi is C^\infty-smooth because Quotient of two smooth functions is smooth. If \varphi(0)=0, then \psi has compact support, because in this case \psi(x)=0 whenever \varphi(x)=0. But if \varphi(0)\ne 0, then \psi  does not have compact support. Indeed, for large enough x we have \varphi(x)=0 and therefore \psi(x)=-\varphi(0)/x. 1 In fact, no, the method of proof is slightly different. You don't need to find a sequence of test functions that converges to zero in C^\infty_c(\Bbb R). Suppose the contrary: there exists a distribution u\in D'(\Bbb R) such that it's restriction on \Bbb R^\ast is represented by \exp(1/x^2). Then, by definition of distribution, we can fix a compact ... 1 Denote by B_m the open ball in R^d of radius m, and let \bar B_m denote the closed ball of radius m. Given a countable collection of non-empty open neighborhoods of f=0, say U_n, fix, for each n, a non-zero function f_n\in U_n such that f_n has support in \bar B_{m_{n+1}}\setminus B_{m_n}, where m_n is a strictly increasing sequence, ... 1 Hint: if a function and all its derivatives have at most polynomial growth, then it defines a tempered distribution. Now check that g(x) satisfies the above hypothesis. 1 To answer this properly, one has to talk about the reason for introducing distributions first. One of the main reasons for this is to be able to define a sort of generalized derivative for objects which are to "singular" to possess derivatives in a classical sense. For example, the functions$$ f\left(x\right)=\begin{cases} 0, & x<0,\\ 1, & ...

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