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I am asked to show that there is a $2\pi$ periodic function $f$ and a $2\pi$ period function $g$ such that $f\in L_{\infty}(\Pi)$ and $g\in \Lambda^{1}(\Pi)$ ($g$ is Lipschitz) with the following properties.

(1) $\displaystyle\lim\limits_{h\rightarrow 0}||f - f(\cdot - h)||_{\infty} \neq 0$ (where $||f||_{\infty} = $ ess sup $\left\{|f(t)|: -\pi \leq t < \pi\right\}$ )

and

(2) $\displaystyle\lim\limits_{h\rightarrow 0}||g - g(\cdot - h)||_{\Lambda^{1}} \neq 0$ (where $||f||_{\Lambda^{1}} = \sup \left\{|f(t)| : -\pi \leq t < \pi\right\} + \sup \left\{\frac{|f(t + y) - f(t)|}{|y|} : -\pi \leq t < \pi, y\neq 0\right\}$ )

Note: By $f(\cdot - h)$ I mean the function which maps $x$ to $f(x-h)$. We use this in class because it is very convenient but I'm not sure if it is standard.

Part (1) is straight forward, I just took $f$ to be the step function $-\chi_{[-\pi,0)} + \chi_{[0,\pi)}$. It turned out that $||f - f(\cdot - h)||_{\infty} = -2$ for all $h > 0$, which was sufficient to establish (1).

I think the reason I am struggling with part (2) so much is that I don't really have a good understanding of what the Lipschitz "metric" is doing. With the $L_{\infty}$ norm the induced metric is simply the largest place where the functions differ. I cannot see an obvious interpretation for the induced metric from the Lipschitz norm. Any advice on where to start?

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It seems that the Lipschitz norm of a funtion $f$ is basically just the sum of the $L_{\infty}$ norm of $f$ and the smallest Lipschitz constant of $f$? For some reason this just doesn't translate easily into metric language for me. –  Kyle Schlitt Sep 30 '11 at 22:06
    
Thank you both. I will play with this further and hopefully I can come up with the proof for it. –  Kyle Schlitt Sep 30 '11 at 22:27
    
I spent almost an hour struggling with this trying to get it to work out. It wasn't until I graphed the function $g - g(\cdot - h)$ for a particular value of $h > 0$ that I finally saw the trick to showing that its Lipschitz norm converges to $2$ as $h\rightarrow 0$. Thank you very much this is a huge breakthrough for me as I am really struggling with this material. :) –  Kyle Schlitt Sep 30 '11 at 23:27

2 Answers 2

up vote 2 down vote accepted

The Lipschitz norm satisfies the following inequality: $$ |g(x)-g(y)|\le\|g\|_{\Lambda_1}|x-y|\quad\forall x,y\;. $$ If $g$ has a bounded derivative, by the mean value theorem $$ |g(x)-g(y)|\le\|g'\|_{\infty}|x-y|\quad\forall x,y\;. $$ We see that in that case, the Lipschitz norm is the $L^\infty$ norm of the first derivative.

This is the intuition you need to solve (2). Can you find a function $g$ such that $g'=f\;$, where $f$ is the example that satisfies (1)?

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complementary answers within $20$ seconds. :-) –  robjohn Sep 30 '11 at 22:13

Consider $g(x)=|x|$ on $[-\pi,\pi)$, the $L^\infty$ part of the norm goes to $0$ since $g$ is continuous. What happens to the smoothness norm?

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