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Chapter 12: Problem 3

Which of the following is a distribution? (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad\left(\lambda_{n} \in \mathbb{R}\right)\) (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad\left(\lambda_{n}, x_{n} \in \mathbb{R}\right)\). (c) \(T(\phi)=(\phi(0))^{2}\). (d) \(T(\phi)=\sup \phi\) (c) \(T(\phi)=\int_{-\infty}^{\infty}|\phi(x)| \mathrm{d} x\).

Short Answer

Expert verified
The distributions are (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad(\lambda_{n}\in \mathbb{R})\) and (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad(\lambda_{n}, x_{n} \in \mathbb{R})\)

Step by step solution

01

Checking Option (a)

Option (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad(\lambda_{n} \in \mathbb{R})\) is a distribution because it is linear in \(\phi\) and also sends compactly supported smooth functions into continuous linear functionals.
02

Checking Option (b)

Option (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad(\lambda_{n}, x_{n} \in \mathbb{R})\) is also a distribution for the same reasons as in Option (a). It is linear in \(\phi\) and sends compactly supported smooth functions into real numbers in a linear way.
03

Checking Option (c)

Option (c) \(T(\phi)=(\phi(0))^{2}\) is not a distribution because it is nonlinear which eventuates into failure to satisfy the required properties of the distributions.
04

Checking Option (d)

Option (d) \(T(\phi)=\sup \phi\) is not a distribution because it is not linear in \(\phi\). The supremum operation is a nonlinear operation, so this does not satisfy the linearity property that distributions must have.
05

Checking Option (e)

Option (e) \(T(\phi)=\int_{-\infty}^{\infty}|\phi(x)| \mathrm{d} x\) is not a distribution because the absolute value operation makes the function nonlinear in \(\phi\), so it does not satisfy the linearity property of distributions

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Most popular questions from this chapter

Show the identities $$ \frac{d}{d x}(\delta(f(x)))=f^{\prime}(x) \delta^{\prime}(f(x)) $$ and $$ \delta(f(x))+f(x) \delta^{\prime}(f(x))=0 $$ Hence show that \(\phi(x, y)=\delta\left(x^{2}-y^{2}\right)\) is a solution of the partial differcntial equation $$ x \frac{\partial \phi}{\partial x}+y \frac{\partial \phi}{\partial y}+2 \phi(x, y)=0 $$

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Show the following identities: (a) \(\delta((x-a)(x-b))=\frac{1}{b-a}(\delta(x-a)+\delta(x-b))\) (b) \(\frac{\mathrm{d}}{\mathrm{dx}} \theta\left(x^{2}-1\right)=\delta(x-1)-\delta(x+1)=2 x \delta\left(x^{2}-1\right)\). (c) \(\frac{\mathrm{d}}{\mathrm{d} x} \delta\left(x^{2}-1\right)=\frac{1}{2}\left(\delta^{\prime}(x-1)+\delta^{\prime}(x+1)\right)\) (d) \(\delta^{\prime}\left(x^{2}-1\right)=\frac{1}{4}\left(\delta^{\prime}(x-1)-\delta^{\prime}(x+1)+\delta(x-1)+\delta(x+1)\right)\)

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