[proofplan] The lower bound $|g(x)|\ge c>0$ first ensures that $g(x)\ne 0$ for every $x\in X$, so the reciprocal function is well-defined. We then estimate the absolute value of the reciprocal pointwise by taking reciprocals of positive [real numbers](/page/Real%20Numbers). This gives the uniform bound $|h(x)|\le 1/c$ for every $x\in X$, and the supremum norm inequality follows directly from the definition of the supremum norm. [/proofplan]

[step:Use the lower bound to define the reciprocal function] Define the function $h:X\to\mathbb R$ by \begin{align*} h(x)=\frac{1}{g(x)} \end{align*} for each $x\in X$. For every $x\in X$, the hypothesis gives $|g(x)|\ge c>0$, and hence $|g(x)|>0$. Therefore $g(x)\ne 0$, so $h(x)$ is a well-defined real number. [/step]

[step:Bound the reciprocal pointwise by taking reciprocals of positive numbers]Fix $x\in X$. Since $|g(x)|\ge c>0$, both $|g(x)|$ and $c$ are positive real numbers. Taking reciprocals reverses the inequality, so \begin{align*} \frac{1}{|g(x)|}\le \frac{1}{c}. \end{align*} Using the elementary identity $|1/a|=1/|a|$ for every nonzero real number $a$, with $a=g(x)$, we obtain \begin{align*} |h(x)|=\left|\frac{1}{g(x)}\right|=\frac{1}{|g(x)|}\le \frac{1}{c}. \end{align*}[/step]

[guided]Fix an arbitrary point $x\in X$. The hypothesis says that the absolute value of $g(x)$ is not merely nonzero, but uniformly separated from zero: \begin{align*} |g(x)|\ge c \end{align*} with $c>0$. Thus $|g(x)|$ and $c$ are positive real numbers. For positive real numbers, the reciprocal map reverses order: if $0<a\le b$, then $1/b\le 1/a$. Applying this with $a=c$ and $b=|g(x)|$ gives \begin{align*} \frac{1}{|g(x)|}\le \frac{1}{c}. \end{align*} Because $g(x)\ne 0$, the reciprocal $1/g(x)$ is defined, and the absolute value of a reciprocal satisfies \begin{align*} \left|\frac{1}{g(x)}\right|=\frac{1}{|g(x)|}. \end{align*} Since $h(x)=1/g(x)$ by definition, we conclude \begin{align*} |h(x)|=\left|\frac{1}{g(x)}\right|=\frac{1}{|g(x)|}\le \frac{1}{c}. \end{align*} The point $x\in X$ was arbitrary, so this estimate holds uniformly for every point of the domain.[/guided]

[step:Take the supremum over the nonempty range of absolute values] The preceding step proves that \begin{align*} |h(x)|\le \frac{1}{c} \end{align*} for every $x\in X$. Since $X$ is nonempty, the set \begin{align*} A=\{|h(x)|:x\in X\} \end{align*} is a nonempty subset of $\mathbb R$. The pointwise estimate says exactly that $1/c$ is an upper bound for $A$. Therefore $h$ is bounded, and by the definition of the supremum norm, \begin{align*} \|h\|_\infty=\sup A=\sup_{x\in X}|h(x)|\le \frac{1}{c}. \end{align*} Since $h=1/g$, this is exactly \begin{align*} \left\|\frac{1}{g}\right\|_\infty\le \frac{1}{c}. \end{align*} [/step]