sorry if this is a very obvious question, but I'm writing a proof for the Kulback-Leibler inequality and the first step is to state $\log(x) \leq x-1$ for all $x>0$.

I get it for $x>1$, but in my notes this isn't stated and when I looked on this site for answers all I could see were proofs showing this was the case for all $x>0$.

As far as I'm aware this demonstrably isn't the case for $0<x<1$ so I'm wondering what I'm missing here.

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    $\begingroup$ Hint: Look at the proof for $x > 1$; repeat it exactly for $0 < x < 1$ (in other words, apply the mean value theorem on the interval $[a, 1]$ where $c$ is any number between $0$ and $1$; apply it to the function $(x-1) - \ln x$. $\endgroup$ – John Hughes Apr 15 at 16:48
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    $\begingroup$ Consider $f(x) = x - 1 - \log(x)$. Show that $f \ge 0$ by showing the minimum of $f$ occurs at $x = 1$. $\endgroup$ – Tom Chen Apr 15 at 16:48

One way is to use calculus.

Define a function $g$ on $(0,\infty)$ by $g(x)= log\ x-x+1$. Now you check that this function attains the local maximum at $x=1$. Therefore $g(x) \leq g(1)$ for every $x \in (0,\infty)$. Hence $\log x \leq x-1$ for all $x \in(0,\infty)$.

  • $\begingroup$ But if $x=0.3$ then $\log(x)=-0.523$ which is greater than $0.3-1=-0.7$? $\endgroup$ – Mark Durkan Apr 15 at 17:04
  • $\begingroup$ @Mark Durkan , By $log$, he means logarithm with natural base. $ln(0.3) = -1.204<-0.7$ $\endgroup$ – Martund Apr 15 at 17:12
  • $\begingroup$ @Martund haha there we go, I knew it was something obvious, cheers! $\endgroup$ – Mark Durkan Apr 15 at 17:22

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