Upper bound on $c \cdot e^x + e^{-cx}$

Question

Is there possible to derive an inequality of the form $e^x + e^{-x} \leq 2 \cdot e^{x^2 / 2}$ (In. 1), but when the left hand side is $c \cdot e^x + e^{-cx}$ and $c \in (0, 1)$, to be upper bounded by some exponential (multiplied by some constant)?

So, in other words, if there is a way to find a constant D and a function f such that: $$ c \cdot e^x + e^{-cx} \leq D e^{f(x, c)} $$

I know that to prove In.1, the proof uses Taylor Expansion for the exponential, but the same trick does not seem to work for the expression $c \cdot e^x + e^{-cx}$.

If you have any suggestion, I would really appreciate. Thank you.

Answer 1

Let set $f(x)=ce^x+e^{-cx}$

• If $x\ge 0$ then $e^{-cx}\le 1$ and $e^x\ge 1$ thus $f(x)\le ce^x+1\le ce^x+e^x\le (c+1)e^x\le 2e^x$

• If $x<0$ then $ce^x<1$ and $e^{-cx}>1$ and similarly $f(x)<2e^{-x}$

So in all cases we get $f(x)\le 2e^{|x|}$

https://www.desmos.com/calculator/2bm34hsicd

Now the inequality you proposed, i.e. $f(x)\le 2\exp(-\frac{x^2}2)$ is actually finer than this rough estimation, but it is still accurate, although a bit more difficult to prove.

Answer 2

Note that $\lim_{|x| \to \infty} {c e^x + e^{-cx} \over e^{{1 \over 2} x^2}} = 0$, hence $x \mapsto {c e^x + e^{-cx} \over e^{{1 \over 2} x^2}}$ is bounded from which the result follows.