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Consider the function $\frac{1}{x}$ defined on the interval $x \in (0, +\infty)$. Is this function strictly positive $f(x) > 0$? It should be obvious but the fact that $\lim\limits_{x\uparrow+\infty}f(x)=0$ makes me uncomfortable. Is it really allowed to say that the function is strictly positive when there exists a limit which is zero?

mathhacker
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    If $x>0$ but $1/x\le 0$, then $x\cdot 1/x=1$ would be either zero or negative, which is not the case. – walkar Oct 26 '22 at 14:16
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    Yes, it is strictly positive because it is positive and non-zero for every $x$ in its domain. $+\infty$ is not part of its domain. – Jaap Scherphuis Oct 26 '22 at 14:16
  • @walkar $f(x)$ is clearly bigger or equals to zero, the question is about the strictness of the inequality – mathhacker Oct 26 '22 at 14:20
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    “Strictly positive” is not the same as “bounded below by a positive number.” Your function is the former but not the latter. – Martin R Oct 26 '22 at 14:24
  • @MartinR thank you, it seems to be correct answer. I am curious what is the formal way to show that? – mathhacker Oct 26 '22 at 14:37
  • What exactly do you want to show formally? A function $f$ is strictly positive if $f(x) > 0$ for all $x$ in the domain. That is the definition of “strictly positive” and you cannot prove it. Or do you want to prove that $1/x > 0$ for all $x > 0$? – Martin R Oct 26 '22 at 14:45
  • @MartinR I want to show formally that "strictly positive" is not the same as "bounded below by a positive number". – mathhacker Oct 26 '22 at 15:26
  • @mathhacker I believe I handled the strict inequality by considering when $1/x \le 0$ -- by contrapositive, if $x>0$ then $1/x>0$. – walkar Oct 26 '22 at 16:12

2 Answers2

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The key point is that a function is only well-defined if it is provided with a domain and codomain, and its properties depend on its behavior strictly on its domain. See (MSE Q1087412) for an overview of what constitutes a function.

Does the function $f:x\mapsto 2x$ take strictly integer values, strictly rational values, all real values, or all complex values? It depends on its domain. Choosing a domain of $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$ could make any of these true. In fact, you could quite convincingly argue that none of these are even the same function, since the domain is a necessary component of the definition.

And the functions $\sin$ and $\cos$ surely cannot be strictly positive? But they might well be if their domains are both limited to $(0,\pi/2)$.

For a slightly more involved example, we may also ask whether functions that are discontinuous on one domain (say, $\mathbb{R}$) might, counterintuitively, be continuous on another (say, $\mathbb{Q}$). And indeed, they can be, for example $\displaystyle g:x\mapsto \begin{cases}1,&x\ge\sqrt{2}\\0,&x<\sqrt{2}\end{cases}$ or the Dirichlet function $\displaystyle h:x\mapsto \begin{cases}1,&x\,\text{rational}\\0,&\,\text{else}\end{cases}$. See (MSE Q1812344) and (MSE Q228233).

With this in mind, the function $i:x\mapsto\frac1x$ is strictly positive over the domain $(0,+\infty)$ despite not being so over $\mathbb{R}/\{0\}$ or, if we permit the use of the extended reals and define $\frac1{+\infty}=0$, the domain $(0,+\infty]$. It doesn't matter how the function behaves outside of its domain.

Jam
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As has been pointed out by @MartinR "strictly positive" is not the same as "bounded below by a positive number". Writing this formally: $\forall{f}$ and $\forall{x} \in \Omega$ and $\forall{a} > 0$ the expression $\neg (f(x) > 0 \iff \inf\{f(\Omega)\} = a)$ is true.

To prove by contradiction, assume that there exist such $f$, $x \in \Omega$ and $a > 0$ such that expression $f(x) > 0 \iff \inf\{f(\Omega)\} = a$ leads to a contradiction. Consider forward implication, $\inf\{f(\Omega)\} = a \iff$ ($\forall{x} \in \Omega$, the inequality $a \leq f(x)$ holds). Choose $f$ and $x$ so that $f(x) = \frac{a}{2} > 0$, then $a \leq \frac{a}{2}$ is a contradiction which concludes the proof.

Considering this, $\frac{1}{x}$ on $(0, +\infty)$ is strictly positive and bounded below by $0$.

mathhacker
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