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If I have a function, say $f(x)$, and I can prove that the function decreases with $x$, and that it converges to a constant when $x = \infty$, does that prove that it decreases at a decreasing rate? In other words, does $f(i+1) < f(i) \text { and } f(\infty) = c \to f(i+1) - f(i+2) < f(i)-f(i+1)$ ? I don't see how it could be any other way, but I often overlook things. Also, would this be the same as saying that, if I can prove the first derivative is negative and that the function converges, that the second derivative must be positive?

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up vote 4 down vote accepted

The rate of decrease will tend to zero in the limit, but that doesn't mean the rate of decrease is itself always decreasing. In other words, $f\to c$ implies $\Delta f\to0$ (the forward difference) and monotonicity ensures $\Delta f<0$, but it's still possible for $\Delta^2 f $ (the second difference) to alternate sign.

Here's a graphic to visualize how it's possible (on a local scale): mspaint

In the above, the black dots represent a monotone decreasing sequence (which we'll say converges to something). We put in between each black term a green term: each green term is slightly lower than the previous black dot, and slightly higher than the next black dot (I put red and blue lines in to make this more apparent), so the green and black dots together create a new monotone decreasing sequence. However, look at the slopes of the orange and purple lines: the slopes get smaller and bigger and smaller and bigger and so on, alternating. But the slopes represent the forward difference $\Delta a_n = a_{n+1}-a_n$, so this means second forward difference $\Delta^2a_n$ is changing sign!

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Wow, thanks so much, the visuals really helped! –  Jand Sep 18 '11 at 4:35
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In other words, does $f(i+1) < f(i) \text { and } f(\infty) = c \to f(i+1) - f(i+2) < f(i)-f(i+1)$?

No. Consider, for example: $$ f(2n) = 5\times2^{-n} $$ $$ f(2n+1) = 3\times2^{-n}$$ This is strictly decreasing $\mathbb N\to \mathbb Q$ and converges towards 0, but the first differences keep oscillating up and down.

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I am sorry but I don't understand your function. What is $f(x)$ in this case? –  Jand Sep 18 '11 at 3:52
    
@JandR He split it up for even and odd cases. –  mixedmath Sep 18 '11 at 3:56
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I'm going to rephrase a bit.

If you have a monotone function that has a finite limit, then you must also have that the limit of the derivative is 0 (and the second derivative too, to address your convexity question).

Why? Assume not - lead to a contradiction. It comes quickly.

I wanted to note that the monotone condition is important here - even a function that does have a limit but is not monotone might have crazy derivatives with highly nonintuitive behavior.

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I think OPs asking if $f$ monotone decreasing implies $f'$ is also monotone, which I believe to be false. –  anon Sep 18 '11 at 3:18
    
Just so I understand, you are saying, "Yes"? Showing $f(i+1)<f(i)$ (for all i) implies monotonicity, and then the fact that the derivative is negative and converges to 0 implies that it decreases at a decreasing rate? –  Jand Sep 18 '11 at 3:21
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$f(x) = \sin^2(\pi x)/x$ satisfies $f(x+1)<f(x)$ for all $x>0$, but is not monotone at all. –  Henning Makholm Sep 18 '11 at 3:25
    
@Henning... good point! hmm... turns out, now that I think about it, the function I am thinking of is discrete, with intervals at 1. Maybe I should open a new question about this. –  Jand Sep 18 '11 at 3:28
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@JandR No, it doesn't. Think of a downward zigzag with a geometric series style scaling factor (so that it converges). The derivative will be getting steeper, less steep, steeper, less steep all the time. Negative, yes. Concave? No. –  mixedmath Sep 18 '11 at 4:12
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