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Is this simple proof for $\ln x \leq x-1$ valid?

Proof:

Since $\ln$ is concave, let $y, x \in \mathbb{R}_{++}.$ We have that $\ln(y) \leq \ln(x) + \frac{d\ln x}{dx}(y-x)$. Since this is valid $\forall x, y \in \mathbb{R}_{++}$, it must be valid for an arbitrary $y$ and $x=1$. Then, we get that $\ln(y) \leq \frac{1}{x}(y-x) \Rightarrow \ln(y) \leq y-1$.

Thanks

Paolo Leonetti
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jpugliese
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  • Consider using \ln to get $\ln$ instead of $ln$ – Fly by Night Mar 15 '18 at 21:12
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    @FlybyNight Better yet, use $\log$ instead of $\ln$. – Mark Viola Mar 15 '18 at 21:19
  • What is you definition of $\log(x)$? For a proof that uses pre-calculus tools only, see THIS ANSWER. – Mark Viola Mar 15 '18 at 21:20
  • @MarkViola There seems to be a UK/US/Applied/Pure divide on this. My experience of UK applied Maths is that $\log$ is often used to denote the base-10 logarithm $\log_{10}$, while $\ln$ is used for the natural logarithm $\ln x \equiv \log_{\mathrm e} x$. (I believe $\ln$ comes from the Latin logarithmus naturalis). Having said that, I've seen $\log$ used for $\log_{\mathrm e}$ a lot in analytic number theory books, even from British authors. – Fly by Night Mar 15 '18 at 21:27
  • Log(x) can be seen as the natural logarithm. Thanks for the pre-calculos @MarkViola but I was just asking if this proof is right. – jpugliese Mar 15 '18 at 21:28
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    @jpugliese Sorry, I hope it is not a stupid question, but from concavity how do you obtain that $\ln(y) \leq \ln(x) + \frac{d\ln x}{dx}(y-x)$? – user Mar 15 '18 at 21:36
  • $\Bbb{R}{++}$ isn't standard notation. Do you mean $\Bbb{R}{>0}$? – Rob Arthan Mar 15 '18 at 21:39
  • @jpugliese Sorry but I can't give an opinion about your proof if you don't explain the first step, I can't really understand where it come from. – user Mar 15 '18 at 22:27
  • @gimusi From the extended MVT, we have $\log(x)=\log(1)+\frac{1}{1}(x-1)+\frac12 \left(-\frac{1}{\xi^2}\right)(x-1)^2$ for some $\xi\in (\min(1,x), \max(1,x))$. The concavity argument simply tells us the sign of the second order term already (it's negative). And this yields the coveted inequality. – Mark Viola Mar 15 '18 at 22:34
  • @MarkViola Ah ok of course! Thanks a lot. Then all works fine! – user Mar 15 '18 at 22:39
  • @jpugliese Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 17 '18 at 23:24
  • Thanks @gimusi. I'm going to do that. The first part of what I wrote follows because, if $f$ is concave and diff, then $f(z) \geq (1-t)f(x) + tf(y)$, where $z = (1-t)x + ty$. We can rewrite z as $z = x + t(y-x)$. Then, $f(x + t(y-x)) - f(x) \geq t(f(y)-f(x))$. Dividing both sides by t, and applying the limit as t tends to zero, we get that the LHS of the above equation will be the $\bigtriangledown f(x) \cdot (y-x)$, i.e., the diretional derivative of $f(x)$ in the direction of the vector $(y-x)$. The result then follows: $\bigtriangledown f(x) \cdot (y-x) \geq f(y) - f(x)$. I applied to ln. – jpugliese Mar 18 '18 at 18:23

2 Answers2

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The proof given in the OP is acceptable given $(i)$ $\log(1)=0$, $(ii)$ $\frac{d\log(x)}{dx}=\frac1x$ and $(iii)$ the logarithm is concave.

Another way forward it to note that since your tacit definition of $\log(x)$ is $\log(x)=\int_1^x \frac1t\,dt$, $x>0$, then we have from elementary analysis of the integral

$$\frac{x-1}{x}=\int_1^x \frac1x \,dt\le \log(x)\le \int_1^x \frac11\,dt=x-1$$

Mark Viola
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    Thanks Mark. My friend just did this proof and was wondering if it was right. Without meaning to be rude, this is a proof-verification post; is the proof I presented correct? – jpugliese Mar 15 '18 at 21:26
  • You're welcome. The proof looks fine given that $\log(1)=0$, $\frac{d\log(x)}{dx}=\frac1x$, and the logarithm function is concave. – Mark Viola Mar 15 '18 at 22:25
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A simply way is to consider $f(x)=x-\ln x -1$ for $x>0$ and note that

  • $f'(x)=1-\frac1x=0 \implies x=1 \quad f(1)=0$
  • $f''(x)=\frac1{x^2}>0$ then $f(1)$ is a minimum
user
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    Thanks gimusi. My friend just did this proof and was wondering if it was right. Without meaning to be rude, this is a proof-verification post; is the proof I presented correct? – jpugliese Mar 15 '18 at 21:25
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    @jpugliese You are right! I check – user Mar 15 '18 at 21:31