Ngô Quốc Anh

March 6, 2010

Convergence in the sense of distribution via an example

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 0:01

Today, let’s talk about the concept of convergence in the sense of distribution. We consider via the following simple example: Assume $u \in C(\mathbb R)$. For each $h \ne 0$, let $u_h$ be the function defined by $\displaystyle {u_h} = \frac{{u(x + h) - u(x)}}{h}$.

Show that $u_h \to u'$ in the sense of distribution as $h\to 0$.

At first, $u$ is assumed to be continuous, therefore $u'$ cannot exist in the classical sense. The point is $u'$ is considered in the sense of distribution theory, that means, $u'$ is a function such that the following holds $\displaystyle\int\limits_\mathbb{R} {u(x)\varphi '(x)dx}= -\int\limits_\mathbb{R} {u'(x)\varphi (x)dx}$

for every test function $\varphi \in C_0^\infty(\mathbb R)$.

Recall that the statement $u_h \to u'$ in the sense of distribution as $h\to 0$ is equivalent to show that $\displaystyle\mathop {\lim }\limits_{h \to 0} \int\limits_\mathbb{R} {\left( {\frac{{u(x + h) - u(x)}}{h}} \right)\varphi (x)dx} = \int\limits_\mathbb{R} {u'(x)\varphi (x)dx}$.

Thus, what we need to prove is the following $\displaystyle\mathop {\lim }\limits_{h \to 0} \int\limits_\mathbb{R} {\left( {\frac{{u(x + h) - u(x)}}{h}} \right)\varphi (x)dx} = - \int\limits_\mathbb{R} {u(x)\varphi '(x)dx}$.

Since $u$ is continuous and $\varphi$ has compact support, it is reasonable to write $\displaystyle\begin{gathered} \int\limits_\mathbb{R} {\left( {\frac{{u(x + h) - u(x)}}{h}} \right)\varphi (x)dx} = \int\limits_\mathbb{R} {\frac{{u(x + h)}}{h}\varphi (x)dx} - \int\limits_\mathbb{R} {\frac{{u(x)}}{h}\varphi (x)dx} \hfill \\ \qquad= \int\limits_\mathbb{R} {\frac{{u(x)}}{h}\varphi (x - h)dx} - \int\limits_\mathbb{R} {\frac{{u(x)}}{h}\varphi (x)dx} \hfill \\ \qquad = \int\limits_\mathbb{R} {\left( {\frac{{\varphi (x - h) - \varphi (x)}}{h}} \right)u(x)dx}. \hfill \\ \end{gathered}$

Furthermore, we can take the limit as $h\to 0$ $\displaystyle\begin{gathered} \mathop {\lim }\limits_{h \to 0} \int\limits_\mathbb{R} {\left( {\frac{{u(x + h) - u(x)}}{h}} \right)\varphi (x)dx} = \mathop {\lim }\limits_{h \to 0} \int\limits_\mathbb{R} {\left( {\frac{{\varphi (x - h) - \varphi (x)}}{h}} \right)u(x)dx} \hfill \\ \qquad= - \int\limits_\mathbb{R} {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\varphi (x - h) - \varphi (x)}}{{ - h}}} \right)u(x)dx} \hfill \\ \qquad= - \int\limits_\mathbb{R} {\varphi '(x)u(x)dx.} \hfill \\ \end{gathered}$

This completes the proof.