What is bounded by $O(T)$ mean?

I have come across something like $$f(T)$$ be bounded by $$O(\sqrt{T})$$ many times in optimization context.

usually, $$f(T)$$ don't have an explicit form.

If $$f(T)$$ do have an explicit form, say $$f_1(T)=\frac{1}{2}\sqrt T$$ and $$f_2(T)=\frac{3}{2}\sqrt T$$, can we say $$f_1(T)$$ and $$f_2(T)$$ are bounded by $$O(\sqrt T)$$?

• Roughly speaking, the graph of $f(x)$ will eventually lie under the graph of $k\sqrt{x}$ for some constant $k$ – Alvin Lepik Aug 17 at 10:32

$$f(x) \in O(g(x))$$ if there exists $$c > 0$$ (e.g., $$c = 1$$) and $$x_0$$ (e.g., $$x_0 = 5$$) such that $$f(x) \le cg(x)$$ whenever $$x \ge x_0$$. You can look this up on Wikipedia ("big $$O$$ notation").
So for $$f_2(T)$$, $$c=3/2$$ and $$g(T)=\sqrt{T}$$.