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.

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