L-1 space of integrable functions

**rogerpodger** · November 4th 2008, 11:47 AM

If f is in L1 and delta > 0, then the L1 norm of f(delta*x)-f(x) goes to zero as delta goes to 1. I have been studying my text for two hours and still really don't understand it. Any thoughts?

By the way, I have some ideas--

If I can show that f is a continuous function of compact support, then I will know that f is dense in L-1. That will guarantee that there exists a function g in L-1 such that the L1 norm of f minus g is less than any epsilon (epsilon greater than zero).

rogerpodger

**Opalg** · November 5th 2008, 01:00 AM

Originally Posted by rogerpodger

If f is in L1 and delta > 0, then the L1 norm of f(delta*x)-f(x) goes to zero as delta goes to 1. I have been studying my text for two hours and still really don't understand it. Any thoughts?

By the way, I have some ideas--

If I can show that f is a continuous function of compact support, then I will know that f is dense in L-1. That will guarantee that there exists a function g in L-1 such that the L1 norm of f minus g is less than any epsilon (epsilon greater than zero).

For $f\in L^1(\mathbb{R})$ and $\delta>0$ , write $M_\delta f(x) = f(\delta x)$ . You want to prove that $\|M_\delta f - f\|\to0$ as $\delta\to1$ . I think that the idea of proving this by doing it first for the case when f is continuous with compact support is the best way to go about it.

The first step is to show that if $f,\,g\in L^1(\mathbb{R})$ then $\|M_\delta f - M_\delta g\| = \delta^{-1}\|f-g\|$ . This is easily seen by making the substitution $y=\delta x$ in the integral $\int_{\mathbb{R}}|f(\delta x) - g(\delta x)|\,dx\ (= \|M_\delta f - M_\delta g\|)$ . That shows that if $\delta$ is close to 1 and $f$ is close to $g$ , then $M_\delta f$ is close to $M_\delta g$ .

The next step is to show that if $g$ is continuous with compact support then $\|M_\delta g - g\|$ is small if $\delta$ is close to 1. This is an easy consequence of the fact that $g$ is uniformly continuous.

Now you just have to put those two steps together and use the triangle inequality: $\|f-M_\delta f\|\leqslant \|f-g\| + \|g-M_\delta g\| + \|M_\delta g - M_\delta f\|$ . If $g$ is close enough to $f$ and $\delta$ is sufficiently close to 1, then all three terms on the right side of the inequality can be made arbitrarily small.

**rogerpodger** · November 5th 2008, 10:08 AM

I was able to solve the problem! Thanks for you help! You gave me somewhere to go without doing the problem for me! I really appreciate it!

rogerpodger

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L-1 space of integrable functions

Thanks!

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