In the following proof, why does the fact that $\displaystyle \lim_{x \to 0} f(x) =0 $ imply that $\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| $ ?

http://www.math.purdue.edu/pow/sprin.../solution9.pdf

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- May 22nd 2010, 10:20 AMRandom Variablelimit proof
In the following proof, why does the fact that $\displaystyle \lim_{x \to 0} f(x) =0 $ imply that $\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| $ ?

http://www.math.purdue.edu/pow/sprin.../solution9.pdf - May 22nd 2010, 11:06 AMmabruka
One way of seeing it:

Absolute value is a continuous function so $\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=\Big|\lim_{x \to 0} f(x) - \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\Big) \Big| =|0-0|=0$

Where it was used the fact that since when$\displaystyle x\to 0 $ then $\displaystyle \frac{x}{2^n}\to 0 $ as well, we have

$\displaystyle \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{\frac{x}{2^n} \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{y \to 0} f(y)=0 $

Finally

$\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=0=\left|\lim_{x \to 0} f(x)\right |=\lim_{x \to 0} |f(x)|$ - May 22nd 2010, 11:08 AMmabruka
- May 22nd 2010, 11:25 AMmabruka
Certainly i dont understand how that step is done,

in fact it would seem to me that $\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)|$ doesnt make any sense since the left side does not depend on x.

Returning to the proof, earlier we had that

$\displaystyle \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|<2\epsilon x$ for all n if $\displaystyle 0< x <\alpha$

Taking $\displaystyle n\to\infty$ and using the continuity of the abs value we get that since if $\displaystyle n\to \infty$ then $\displaystyle y=\frac{x}{2^n}\to 0$,

in our case :

if $\displaystyle n\to \infty $ then $\displaystyle f(\frac{x}{2^n})=f(y)\to 0$.

And we end up having $\displaystyle |f(x)|\leq 2\epsilon x $ from where the proof follows quite easily. - May 22nd 2010, 11:30 AMRandom Variable
- May 22nd 2010, 12:08 PMmabruka
Either a typo or some argument (which is not clear, at least for me) equivalent to the n-limit one. So far i dont see any flaws on this one.

- May 23rd 2010, 02:39 AMHallsofIvy
It certainly looks to me like this was supposed to be

$\displaystyle \lim_{n\to \infty}\left|f(x)- f(x/2^n)\right|= |f(x)|$

Because the right hand side depends on x, not n. But, since that is $\displaystyle \left|f(x)- \lim_{n\to\infty}f(x/2^n)\right|$ it looks obvious.