[proofplan] We prove both directions by unpacking the definition that a real-valued function is bounded on its domain precisely when its image is a bounded subset of $\mathbb{R}$. In the forward direction, boundedness of $f(X)$ gives upper and lower bounds for all values of $f$, and a single positive constant is chosen to dominate both bounds in absolute value. In the reverse direction, the assumed absolute-value estimate places the whole image inside the interval $[-M,M]$, which is a bounded subset of $\mathbb{R}$. [/proofplan]

[step:Convert boundedness of the image into one absolute bound]Assume that $f$ is bounded on $X$. By definition, the image $f(X):=\{f(x):x\in X\}$ is a bounded subset of $\mathbb{R}$. Hence there exist [real numbers](/page/Real%20Numbers) $a,b\in\mathbb{R}$ such that \begin{align*}a\le y\le b \qquad \text{for every } y\in f(X)\end{align*} Define $M:=\max\{|a|,|b|,1\}$. Then $M>0$. For every $x\in X$, the value $f(x)$ belongs to $f(X)$, so \begin{align*}a\le f(x)\le b\end{align*} Since $a\ge -|a|\ge -M$ and $b\le |b|\le M$, it follows that \begin{align*}-M\le f(x)\le M\end{align*} Therefore \begin{align*}|f(x)|\le M \qquad \text{for every } x\in X\end{align*}[/step]

[guided]Assume that $f$ is bounded on $X$. The definition of boundedness for a real-valued function says that the set of all values of the function is bounded in $\mathbb{R}$. We denote this image by $f(X):=\{f(x):x\in X\}$. Thus there are real numbers $a,b\in\mathbb{R}$ such that every value in the image lies between them: \begin{align*}a\le y\le b \qquad \text{for every } y\in f(X)\end{align*} The desired estimate uses one positive number that controls both positive and negative values of $f$. To obtain such a number, define \begin{align*}M:=\max\{|a|,|b|,1\}\end{align*} The inclusion of $1$ ensures $M>0$, even in degenerate cases where the other quantities are zero. Now let $x\in X$ be arbitrary. Since $f(x)\in f(X)$, the bounds on the image give \begin{align*}a\le f(x)\le b\end{align*} Because $|a|\le M$, we have $-M\le -|a|\le a$. Because $|b|\le M$, we have $b\le |b|\le M$. Combining these inequalities yields \begin{align*}-M\le f(x)\le M\end{align*} For a real number $t$, the inequality $-M\le t\le M$ is equivalent to $|t|\le M$ when $M\ge 0$. Applying this with $t=f(x)$ gives \begin{align*}|f(x)|\le M\end{align*} Since $x\in X$ was arbitrary, the estimate holds for every $x\in X$.[/guided]

[step:Use the absolute bound to prove boundedness of the image] Conversely, assume that there exists $M>0$ such that \begin{align*}|f(x)|\le M \qquad \text{for every } x\in X\end{align*} For every $y\in f(X)$, there exists $x\in X$ such that $y=f(x)$. Hence \begin{align*}|y|=|f(x)|\le M\end{align*} Therefore \begin{align*}-M\le y\le M \qquad \text{for every } y\in f(X)\end{align*} Thus $f(X)$ is a bounded subset of $\mathbb{R}$. By the definition of boundedness for real-valued functions, $f$ is bounded on $X$. This proves the equivalence. [/step]