I have a function $h(x):K\rightarrow\mathbb{R}$, where $K\subset\mathbb{R}$ is compact. The function itself is uniformly continuous, and therefore bounded. Moreover, the function is nonzero over $K$. I would like to show that the magnitude of reciprocal of the function, $\big|\frac{1}{h(x)}\big|$ is also bounded on $K$. I am convinced all of the pieces are here, but I'm at a loss as to how to put them together. I would love a hint/response!
Proving that the reciprocal of a function is bounded
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calculus
real-analysis
3 Answers
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Because $h$ is continuous on $K,$ so is $|h|.$ Hence $|h|$ assumes a minimum value $m$ at some $x_0\in K.$ Because $|h(x_0)|>0,$ we have $m> 0.$ But notice $|h| \ge m \implies \frac{1}{|h|} \le \frac{1}{m} ,$ and we're done.
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Since $h$ is bounded, you know that there exists $k_1,k_2$ such that
$$\forall x\in K,\quad k_1\leq \vert h(x)\vert\leq k_2.$$
And since the function is non-zero, $[k_1,k_2]\subset (0,+\infty)$ so
$$\forall x\in K,\quad \frac 1{k_2}\leq \vert \frac 1{h(x)}\vert\leq \frac 1{k_1}$$
because $x\mapsto \frac 1x$ is decreasing.
You can then conclude that $x\mapsto \vert \frac 1{h(x)}\vert $ is also bounded.
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If it is unbounded, then there is a sequence $(x_n) \subset K$ such that $\pm \infty$ is a cluster point of $1/f(x_n)$. Hence, $0$ is a cluster point of $f(x_n)$, so $(x_n)$ has a subsequence $(y_n)$ such that $f(y_n) \to 0$. By the compactness of $K$, $(y_n)$ has a convergent to some $z \in K$ subsequence $(z_n)$, and by the continuity of $f$, $f(z) = \lim f(z_n) = 0$, $(f(z_n))$ being a subsequence of $(f(y_n))$. A contradiction.