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Prove that if $f:R^+ \rightarrow R^+$ is continuous on the positive reals and is decreasing, then for all $a$ there exists an $\eta > 0$ such that $(a-\eta)f(a-\eta) > \frac{1}{2}a*f(a)$.

EDIT - The proof in the error was identified, and I can see how it can be easily proved just by using the fact that it is decreasing. However, is it possible to prove this statement just using the continuity condition?

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  • $\begingroup$ There is a mistake in your "Because $f$ is decreasing" sentence, the inequality involving $f(x)$ is incorrect. $\endgroup$ – process91 Jun 26 '14 at 20:51
  • $\begingroup$ EDIT - I see the error $\endgroup$ – kritkally Jun 26 '14 at 20:55
  • $\begingroup$ It seems to me that it is possible to prove this without continuity - just use the decreasing property. $\endgroup$ – process91 Jun 26 '14 at 20:55
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Here's a hint on how to prove it: Ignore continuity. You have the decreasing property, so you know that for $0<\eta<a$, you have that $$f(a-\eta)>f(a).$$

You want to get $\frac a 2$ on the right, and you could certainly do that if you multiplied everything by $\frac a 2$, so how can you tweak this to end up with $\frac a 2$ on the right and $(a-\eta)$ on the left? (Hint: restrict your choice of $\eta$ further.)

You can also completely ignore the decreasing property. The fact that this might be possible should be intuitively signaled by the fact that $\lim_{\eta \to 0}(a-\eta)f(a-\eta) = af(a)>\frac 1 2 a f(a)$ if $f$ is continuous at $a$. We should turn this visual picture into a proof as follows:

Since $f$ is continuous, there exists some $\delta>0$ such that $|(a-\eta)-a|=|\eta|<\delta$ implies $|(a-\eta)f(a-\eta)-af(a)|<\frac 1 2 a f(a)$. Therefore $$-\frac 1 2 a f(a) < (a-\eta)f(a-\eta)-af(a) < \frac 1 2 a f(a)$$ but we don't really care about the right side, because adding $af(a)$ to the left and the middle yields $$\frac 1 2 a f(a)<(a-\eta)f(a-\eta).$$ Literally having a visual picture to look at helps tremendously. This shows that not only is there at least one $\eta$ that works, but a whole interval of them.

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  • $\begingroup$ I understand this proof clearly, but is it possible to prove the statement without the fact that it is decreasing? $\endgroup$ – kritkally Jun 26 '14 at 21:04
  • $\begingroup$ It is possible, I will post a proof now. $\endgroup$ – process91 Jun 26 '14 at 21:35

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