# Proving that 2 functions are bounded.

Let $$f,g:\mathbb R \to \mathbb R$$. Let $$h_1$$ = $$f-g$$ and $$h_2$$ = $$f+g$$. Suppose $$h_1$$ and $$h_2$$ are bounded. Show that $$f$$ and $$g$$ are bounded.

I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.

• Express $f,g$ in the form of $h_2, h_1$. – xbh Dec 18 '18 at 4:57
• $|f(x) \pm g(x) | \le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded. – Henno Brandsma Dec 18 '18 at 5:11

The sum of bounded functions is bounded and so are scalar multiples. Note that $$f= \frac{h_1 + h_2}{2}$$.
Likewise for differences of bounded functions and $$g = \frac{h_2 - h_1}{2}$$.
Suppose $$h_1, h_2$$ are bounded. Let $$D(f)$$ denote the domain of $$f$$, where $$f$$ is a function. Then $$\exists M_1, M_2$$ such that $$h_1(x) = f(x) - g(x) \leq M_1 \forall x \in D(h_1),$$ $$h_2(x) = f(x) + g(x) \leq M_2 \forall x \in D(h_2).$$ So, we have that for arbitrary $$x \in D(f) \cap D(g)$$, $$f(x) - g(x) + f(x) + g(x) = 2f(x) \leq M_1 + M_2 \implies f(x) \leq \frac{M_1 + M_2}{2},$$ so $$f$$ is bounded, by definition. Similarly, notice that $$g(x) - f(x) + g(x) + f(x) = 2g(x) \leq M_2 - M_1 \implies g(x) \leq \frac{M_2 - M_1}{2},$$ so $$g$$ is bounded, by definition. This completes the proof.
• Maybe it should be $x\in D(h_1)\cap D(h_2)$? – Shubham Johri Dec 18 '18 at 5:54