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The tittle says it all. I think it's true, and I tried to prove it by showing that the derivative of this function: $-2Bxe^{-Bx^2}$ is bounded from above with a bound less than 1, in order to do that, I tried to use Taylor series of $e^{-Bx^2}$, but it seems that leads nowhere. Any suggestion?

Here $B>0$ is a real number and we consider the euclidean norm.

Daniel
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2 Answers2

1

You want to check whether the maximum of $|f'|$ is less than $1$. So take one more derivative:

$$f''(x)=-2Be^{-Bx^2}+4B^2x^2e^{-Bx^2}.$$

This is zero if and only if $4B^2x^2-2B=0$, i.e. $x^2=1/(2B)$. At these points you have $|f'(x)|=2^{1/2} B^{1/2} e^{-1/2}$. You can check that these must be the points where $|f'|$ is largest since $f'(0)=0$ and $f'(x) \to 0$ as $x \to \pm \infty$. It is clear that this grows without bound as a function of $B$, so $e^{-Bx^2}$ cannot be a contraction for all $B$.

Ian
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  • You are wrong on the second differentiation and also why do you even look on the second derivative? – mathreadler Mar 09 '15 at 18:55
  • @mathreadler I am attempting to maximize $|f'|$. Also I had a typo in the first line that I had corrected in the rest of the exposition. – Ian Mar 09 '15 at 18:55
  • I think he means contraction as in |f(x)| < |x|, i.e. operator norm $\leq 1$ – mathreadler Mar 09 '15 at 18:58
  • Mathreadler, I don't mean that when I say contraction. I talked about that in a previous comment. – Daniel Mar 09 '15 at 19:00
  • @Ian I understand your point. The problem is that I need this result for ALL $B>0$. I must say that maybe what I've claimed is not true (that's why I've said "Is the function... a contraction?") – Daniel Mar 09 '15 at 19:01
  • @Devilathor I didn't finish the problem. The maximum of $|f'|$ is $2^{3/2} B^{3/2} e^{-2B^2}$. You now need to maximize this and show that it is less than $1$. This problem is of a similar character to the previous optimization problem. – Ian Mar 09 '15 at 19:02
  • I'm trying to solve an exercise about roots of the function $x+e^{-Bx^2}cos(x)$ so I translated that into the language of fixed point. In order to use the Bannach Fixed Point Theorem, It suffices to show that $e^{-Bx^2}$ is a contraction – Daniel Mar 09 '15 at 19:03
  • @Ian. Wouldn't this process lead to an "infinite" chain of optimizations? – Daniel Mar 09 '15 at 19:13
  • @Devilathor No, once you've optimized over both variables you have just a number, not a function. Wolfram Alpha tells me this number happens to be $\left ( \frac{3}{2e} \right )^{3/4} \approx 0.64$, and is attained at $B=\frac{\sqrt{\frac{3}{2}}}{2} \approx 0.61$. – Ian Mar 09 '15 at 19:20
  • @Ian. Of course. I don't know what was I thinking. I'm trying your suggestion but I couldn't maximize that function by hand. – Daniel Mar 09 '15 at 19:22
  • @Ian. So, according that data, the function is a contraction indeed – Daniel Mar 09 '15 at 19:23
  • @Devilathor Correct. Note that your problem is harder than you might think. $x+e^{-Bx^2} \cos(x)=0$ if and only if $-e^{-Bx^2} \cos(x)=x$, so you actually need to show that $-e^{-Bx^2} \cos(x)$ is a contraction. And this function is actually not always a contraction, so your problem is somewhat more difficult. (Cf. http://www.wolframalpha.com/input/?i=maximum+of+d%2Fdx+e^%28-x^2%29+cos%28x%29 ) You can still use the Banach Fixed Point Theorem in this context, but you need to choose an appropriate compact subset of $\mathbb{R}$ where your function really is a contraction ($[0,1]$ is fine). – Ian Mar 09 '15 at 19:26
  • @Ian I noted that, but $-e^{-Bx^2}\mbox{cos}(x)$ will be a contraction if $-e^{-Bx^2}$ is, isn't that right? – Daniel Mar 09 '15 at 19:31
  • @Devilathor Not globally. The first product rule term, where the derivative hits the exponential, will be less than $1$ in magnitude. But the other term, where the derivative hits the cosine, may be big enough to push the whole thing larger than $1$ in magnitude. My Wolfram Alpha link above shows that this actually does happen when $B=1$ at $x \approx -0.6$. – Ian Mar 09 '15 at 19:33
  • I thought in restricting the domain too, but the exercise asks to prove that there is ONLY ONE root over all reals. I think that all this discussion shows that using Bannach Fixed Point Theorem is not the way to attack this problem. However, that's what occurred to me and I can't see any other way to solve it. – Daniel Mar 09 '15 at 19:35
  • @Devilathor Ultimately it boils down to the Banach fixed point theorem, but you need to be sure that there aren't any roots outside your compact set. But this isn't too bad. There can't be any roots outside $[-1,1]$, because $|e^{-Bx^2} \cos(x)| \leq 1$. There can't be any roots in $[-1,0]$ either, because $e^{-Bx^2} \cos(x)>0$ on $[-1,1]$. So if there is a root, it is in $[0,1]$. Now you need to shrink the domain a little bit more (because the derivative is too big near x=0.59) to get it to be a contraction. – Ian Mar 09 '15 at 19:37
  • @Devilathor You also, unfortunately, must show that it maps the restricted domain into itself, which will probably not be a trivial matter. – Ian Mar 09 '15 at 19:38
1

Suppose $f$ is a contraction with rank $\lambda <1$. In particular, we must have $|f'(x)| \le \lambda$ for all $x$, or equivalently ${2B|x| \over e^{B x^2}} \le \lambda$ for all $x$.

If we pick $B> ({e \over 2})^2$, then $f'(-{1 \over \sqrt{B}}) = { 2 \sqrt{B} \over e} > 1$.

More specifically, let $B=e^2$, then $f'(-{1 \over e}) = 2$.

copper.hat
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  • So your point is that the derivative cannot be bounded by 1 for all $B>0$? – Daniel Mar 09 '15 at 19:38
  • @Devilathor: Yes, for sufficiently large $B$, $f$ is not a contraction. To be more specific, if you choose $B=e^2$, then $f'(-e^{-1}) = 2$. – copper.hat Mar 09 '15 at 19:39
  • Did you read the discussion of Ian's answer? There, it seems that what you say is not true... – Daniel Mar 09 '15 at 19:40
  • Well, just do the computation in my previous comment. – copper.hat Mar 09 '15 at 19:40
  • ...Hm, one of us has made an error, and I am not sure which... – Ian Mar 09 '15 at 19:41
  • I didn't want to offend. I did the computations and indeed, what you say is true. So yes. there's a mistake around... – Daniel Mar 09 '15 at 19:42
  • No, I think I did, I'm just not sure what it is. At any rate, Devilathor has actually reduced his original problem to a problem which is not equivalent to his original problem. I would propose that Devilathor open another question to deal with his original problem about finding roots of $x+e^{-Bx^2} \cos(x)$, i.e. fixed points of $-e^{-Bx^2} \cos(x)$. – Ian Mar 09 '15 at 19:44
  • Indeed it seems I did make a mistake: http://www.wolframalpha.com/input/?i=maximum+of+2xye^%28-yx^2%29%2Cy%3E0 My technique works, I think I just made an algebraic mistake. I've edited my answer to remove the part where it says it is a contraction. My comments on my answer will remain problematic... – Ian Mar 09 '15 at 19:46
  • @Devilathor: I think my computation is correct. – copper.hat Mar 09 '15 at 19:46
  • I found my mistake: I turned $4B^2x^2-2B=0$ into $x^2=2B$ instead of $x^2=1/(2B)$. – Ian Mar 09 '15 at 19:50
  • I posted my original problem as a question by itself: http://math.stackexchange.com/questions/1182748/show-that-xe-bx2-mboxcosx-has-only-one-root-over-all-reals-b0 I hope some of you can help me. Ian, your answer was very important, of course, you get the accepted one! Thank you very much. – Daniel Mar 09 '15 at 19:53
  • @Ian: 2B or not 2B? – copper.hat Mar 09 '15 at 19:53
  • @copper.hat Nice one!! – Daniel Mar 09 '15 at 19:55
  • Hmmm, strange the way this works... – copper.hat Mar 09 '15 at 20:03