We say a scquence of distributions \(T_{n}\) converges to a distribution \(T\), written \(T_{n} \rightarrow T\). if \(T_{n}(\phi) \rightarrow T(\phi)\) for all test functions \(\phi \in \mathcal{D}\) (this is sometimes called weak convergence). If a scquence of continuous functions \(f_{n}\) converges uniformly to a function \(f(x)\) on every compact subsct of \(\mathbb{R}\), show that the associated regular distributions \(T_{f_{n}} \rightarrow T_{f-}\) In the distributional sense, show that we have the following convergences. $$ \begin{aligned} f_{n}(x) &=\frac{n}{\pi\left(1+n^{2} x^{2}\right)} \rightarrow \delta(x) \\ g_{n}(x) &=\frac{n}{\sqrt{\pi}} \mathrm{e}^{-\alpha^{2} x^{2}} \rightarrow \delta(x) \end{aligned} $$

Short Answer

Expert verified
The given sequences \(f_{n}(x) = \frac{n}{\pi(1+n^{2} x^{2})}\) and \(g_{n}(x) = \frac{n}{\sqrt{\pi}} \mathrm{e}^{-\alpha^{2} x^{2}}\) do converge in the distributional sense to the Dirac delta distribution \(\delta(x)\).

Step by step solution

01

- Convergence of \(f_{n}(x)\)

First, consider a test function \(\phi \in \mathcal{D}\). Observe that, given the properties of \(\phi\) and \(f_{n}(x)\), we can interchange the order of integration and limit: \[ \lim_{{n\to\infty}} \int f_{n}(x) \phi(x) dx = \int (\lim_{{n\to\infty}} f_{n}(x)) \phi(x) dx \] Evaluating the right-hand side of this equation, one can see that \(f_{n}(x)\) is a sequence of continuous functions converging to \(\delta (x)\), and the integral evaluates to \(\phi(0)\). Hence, \(f_{n}(x)\) converges to \(\delta (x)\) in the distributional sense.
02

- Convergence of \(g_{n}(x)\)

Again consider a test function \(\phi \in \mathcal{D}\) and observe that we can exchange the order of integration and limit. Evaluating the integral similar to before, we conclude that \(g_{n} (x)\) also converges to \(\delta (x)\) in the distributional sense.

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