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Let $x\in \mathbb{R}$ and consider the function $f(x)=\log(e+e^{x})$ and note that $f(x)\geq 1$. I am looking for an "good" upper bound for $f$. Clearly one has the estimate $$\log(e+e^{x})\leq e+|x|,$$ but I am looking for an estimate such that $f$ stays bounded for $x\to -\infty$.

Has someone got any ideas?

4 Answers4

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Let $f(x)=\ln(e+e^x)$. We can rewrite $f(x)$ as $f(x)=\ln \left (e^{1+x \over 2} [e^{-{1-x \over 2}}+e^{{1-x \over 2}}]\right )$. Thus $f(x)={1+x\over 2} + \ln [e^{-{1-x \over 2}}+e^{{1-x \over 2}}]$.

Now we note that $\ln [e^{-{1-x \over 2}}+e^{{1-x \over 2}}]=\ln(2) + \ln [0.5e^{-{1-x \over 2}}+0.5e^{{1-x \over 2}}]=\ln(2) + \ln[\cosh({1-x \over 2})]$. Since $\ln [0.5e^{-{1-x \over 2}}+0.5e^{{1-x \over 2}}]$ is midpoint-concave i.e. $\ln [0.5e^{-{1-x \over 2}}+0.5e^{{1-x \over 2}}] \ge -0.5{1-x \over 2}+0.5{1-x \over 2}=0$, we have $$ f(x) \ge \max(1,{1+x\over 2} + \ln(2)). $$

Now since $\cosh(x) \le e^{x^2 \over 2}$ we have $$ f(x) \le {1+x\over 2} + \ln (2) +{[{1-x \over 2}]^2 \over 2}. $$ (corrected bound due to the @Gary comment).

By the way, the inequality for $\cosh$ appeared some time ago on math.stackexchange (cosh x inequality).

The comment of @Yves Daoust forced me to think of another upper bound for all $x$. Here it is: $$ f(x) \le 1+ {e^{x-1} \over \sqrt{1+e^{x-1}}}. $$

rrv
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    The lower bound is unbounded for $x\to-\infty$, and the upper bound is not defined for $y<x$ !? –  Jun 10 '20 at 10:04
  • My apologies. Made too many mistakes in the first answer attempt. Seems all are corrected now. – rrv Jun 10 '20 at 11:57
  • Hem, what is the benefit of an upper bound that is as complicated than the original function ? –  Jun 10 '20 at 11:59
  • @YvesDaoust You are right. There is no benefit of a bound, which is more complex than the original function. This is just what i can. Surely someone can do better and your answer below looks better. – rrv Jun 10 '20 at 12:08
  • You missed the $\log$ when arriving at $\cosh$ for the first time. This is the reson your upper bound grows exponentially (which does not make much sese). – Gary Jun 10 '20 at 13:32
  • Yes indeed. @Gary thank you. Corrected. – rrv Jun 10 '20 at 14:32
  • It probably modifies your very last bound as well. – Gary Jun 10 '20 at 15:10
  • The last bound is not based on the preceeding derivations and thus no modification is needed. According to me it is correct and quite tight for $x<0$. Anyway it is easy to give it a quick check – rrv Jun 11 '20 at 06:11
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$$\begin{cases}x\le a\to\ln(e+e^a),\\a\le x\le b\to\dfrac{\ln(e+e^b)-\ln(e+e^a)}{b-a}(x-a)+\ln(e+e^a),\\x\ge b\to x-b+\ln(e+e^b)\end{cases}$$

You choose $a<b$ freely. This is piecewise linear.

Alternatively,

$$\frac{x+\sqrt{x^2-2x+2}}2+\ln(2).$$

enter image description here

This is reasonably tight, with asymptotes parallel to those of the original function. And more generally,

$$\frac{x+\sqrt{(x-a)^2+1}-a-1}2+\ln(e+e^a).$$

Community
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Does $$ \log(e+e^x)\leq \log(1+e)+\max\{0,x\} $$ suffice? The bound gets worse as $x\to\infty$ but it's bounded

whpowell96
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Using this bound New bound for Am-Gm of 2 variables we get :

$$\ln(e+e^x)\leq \ln\Big(\sqrt{e^{x+1}}+\Big(e^ee^{xe^x}\Big)^{\frac{1}{e+e^x}}\Big)$$

It's the sharper I can get .

Miss and Mister cassoulet char
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