# About the concept bound a function

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

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

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?

• Can you elaborate on point $(2)$? Why don't you think such an $M$ exists? (Just take $M=1+\vert c\vert$.) – Noah Schweber Jul 8 at 21:49

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 $$c<0$$.
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 $$101$$.