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From wikipedia:

In mathematics, a function f defined on some set X with real or complex values is called bounded if the set of its values is bounded. In other words, there exists a real number M such that

$|f(x)| \leq M, \forall x \in X.$

I have a series of conceptual doubts about this:

  1. Does strict inequality need to be considered bounded? In wikipedia says that $\arctan(x)$ is bounded since $|\arctan(x)|< \frac{\pi}{2}$, but i want to confirm with mathematicians of MSE.

  2. The concept of bounded seems symmetric, i mean, what will happen to functions like $f(x) = \sin(x) + c$, where $c-1 \leq f(x)\leq c+1$? in this case, i cant say that there exist an $M$ such that $|f(x)| \leq M$, so, this function is considered bounded?

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  • $\begingroup$ Can you elaborate on point $(2)$? Why don't you think such an $M$ exists? (Just take $M=1+\vert c\vert$.) $\endgroup$ – Noah Schweber Jul 8 at 21:49
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  1. If $|\arctan(x)|<\frac{\pi}{2}$ then certainly we can also write $|\arctan(x)| \le \frac{\pi}{2}$, so yes we say that $\arctan$ is bounded.

  2. If $c>0$ and $c-1 \leq f(x)\leq c+1$, then it's also true that $-c-1 \le f(x) \le c+1$, in which case we get $|f(x)|\le c+1$, meaning $f$ is bounded. A similar argument applies if $c<0$.

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There is an $M$ such that the weak inequality holds if and only if there is an $M$ for which the strong inequality holds, so you do not have to worry about which inequality you use. They define the same concept.

If $f$ is bounded then for any constant $c$, $f+c$ is bounded. For example, $\sin$ is bounded by $1$ (also by $200$). So $100 + \sin$ is bounded by $101$.

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