Max vs Integral

Recall that the fundamental theorem of calculus states

$f(b) - f(a) = \int_a^b f' (x) \, dx$

Thus, for all $a, b \in [0,1]$ , we have

$| f(b) | - | f(a) | \leq | f(b) - f(a) | = \left| \int_a^b f'(x) \, dx \right| \leq \int_a^b \left| f'(x) \right| \, dx \leq \int_0^1 \left| f'(x) \right| \, dx$

Now, let $b$ be the x-value that maximizes $|f(x)|$ , and $a$ be the x-value that minimizes $|f(x)|$ . Then, we have

$|f(a)| = \min |f(x)| \leq \int_0^1 |f(x) | \, dx$

Thus

$\max | f(x) | = |f(b)| \leq |f(a) | + \int_0^1 \left| f'(x) \right| \, dx = \int_0^1 \left| f'(x) \right| + \left| f(x) \right| \, dx$

---

We have inequality whenever any of the inequalities do not hold. E.g. if $f(x)$ has both positive and negative values, or if $f'(x)$ has both positive and negative values, of when $\{ a, b \} \neq \{ 0, 1 \}$ along with non-zero $f'(x), f(x)$ outside of $[a,b]$ .

Suppose that $m = \inf_{x \in [0,1]} f(x)$ and $M = \sup_{x \in [0,1]}f(x)$ . Find $c,d \in [0,1]$ such that $f(c) = m$ and $f(d) = M$ .

Suppose first that $f$ is non-negative throughout $[0,1]$ . Then $0 \le m \le M$ and $A=M$ . Since $f(x) \ge m$ for all $x \in [0,1]$ , it follows that $m \; = \; \int_0^1 m\,dx \; \le \; \int_0^1 f(x)\,dx \; = \; \int_0^1 |f(x)|\,dx$ Also $M-m \; = \; \left|\int_c^d f'(x)\,dx\right| \; \le \; \left|\int_c^d |f'(x)|\,dx \right| \; \le \; \int_0^1 |f'(x)|\,dx$ Adding these results tells us that $M \le B$ , and so $A \le B$ in this case. A similar argument shows that $A \le B$ when $f$ is non-positive throughout $[0,1]$ .

Suppose now that $f$ can be both positive and negative. Thus $m < 0 < M$ and $A = \mathrm{max}(|m|,M)$ . Find $u \in [0,1]$ such that $f(u) = 0$ . Then $|m| \; = \; \left|\int_u^c f'(x)\,dx\right| \; \le \; \int_0^1 |f'(x)|\,dx \; \le \; B \hspace{2cm} M \; = \; \left|\int_u^d f'(x)\,dx\right| \; \le \; \int_0^1 |f'(x)|\,dx \; \le \; B$ and hence $A = \mathrm{max}(|m|,|M|) \le B$ in this case as well.