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.

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

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    $\begingroup$ Maybe it should be $x\in D(h_1)\cap D(h_2)$? $\endgroup$ – Shubham Johri Dec 18 '18 at 5:54
  • $\begingroup$ Ah, you are right; good catch! I've edited my answer to include this. $\endgroup$ – alexsieusahai Dec 18 '18 at 5:58

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