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$$f: \mathbb R \rightarrow \mathbb R \quad with \quad f(\theta) =\begin{cases} \frac{1}{\theta^2}log\left\{ \frac{e^{\theta} + e^{-\theta}}{2} \right\}, \theta \neq 0 \\ 1/2, \theta = 0 \end{cases}.$$

My attempts:

1rst: I tried to proove f is differentiable at $x=0,$ but the computations grew really fast. My plan was to show the derivative at zero is zero;

2nd: I differentiate $f$ for $\theta \neq 0.$ I was looking forward proving it is decreasing for $\theta > 0$ and increasing for $\theta < 0.$ But I couldn't deal with the inequalities that emerged;

3rd: I tried to use a reductio ad absurdum argument, but I couldn't conclude the absurd I was searching for;

4th: since a injective continuous function must be monotone, I came up with the idea of proving $f$ is injective in $(0, \infty)$ and $(-\infty,0).$ But the expressions wered really tough to struggle with.

Pagginelli
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    Use the definition of bounded and show that the function satisfies the definition, or at least attempt to show what you have tried. – jimjim Dec 26 '17 at 01:43

2 Answers2

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Hint:

(written before the question was modified to explicitly ask for a proof that the upper bound of $f$ is $\frac{1}{2}$)

  • Show that, for $\theta\ge 1$, we have $0\le f(\theta)\le \frac{1}{\theta^2}\log e^\theta=\frac{1}{\theta}\le 1$.
  • $f$ is an even function, so similarly, for $\theta \le -1$, we have $0\le f(\theta)\le 1$.
  • Prove that $\lim_{\theta\to 0}f(\theta)=f(0)$, which means that $f$ is continuous in $\theta=0$. (For example, use L'Hospital's rule twice.) In other points $\theta\ne 0$ it is continuous as a composition of continuous functions. That will make $f$ continuous on the whole $\mathbb R$, and in particular on $[-1, 1]$. Due to the Extreme value theorem, it is then bounded on $[-1, 1]$.
  • Being bounded on $(-\infty, -1]$, $[-1, 1]$, and $[1, +\infty)$, the function is bounded on the whole of $\mathbb R$.
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Please see cosh x inequality for proofs that $\cosh x =\frac{e^x+e^{-x}}{2}\le e^\frac{x^2}{2}$. By taking $\log$ of both sides, and then dividing by $x^2$, we get: $\log\left\{\frac{e^x+e^{-x}}{2}\right\}\le \frac{x^2}{2}$, or $\frac{1}{x^2}\log\left\{\frac{e^x+e^{-x}}{2}\right\}\le\frac{1}{2}$ for every $x\in\mathbb R, x\ne 0$. The case $x=0$ is trivial.