Average value of complex valued function property

Question

If $f:[a,b] \to \mathbb{C}$ is a continuous function define the average of $f$

$$A = \frac{1}{b-a} \int_a^b f(x)dx$$ where $\int f= \text{Re} (f) + i \int \text{Im}(f)$

Then I need to show that if $|f| \le |A|$ on $[a,b]$ then $f=A$. How to approach this?

Answer 1

Assume that $A\ne 0$. We have \begin{align*} |A|&=\dfrac{1}{b-a}\left|\int_{a}^{b}f(x)dx\right|\leq\dfrac{1}{b-a}\int_{a}^{b}|f(x)|dx\leq\dfrac{1}{b-a}\int_{a}^{b}|A|dx=|A|, \end{align*} so \begin{align*} \dfrac{1}{b-a}\int_{a}^{b}(|A|-|f(x)|)dx=0, \end{align*} but $|A|-|f(\cdot)|\geq 0$ and $|A|-|f|$ is continuous, one has $|f|=A$.

Now we let $z$ be a complex number such that \begin{align*} zA=z\cdot\dfrac{1}{b-a}\int_{a}^{b}f(x)dx\geq 0, \end{align*} by writing $zf=u+iv$ for real $u,v$, then \begin{align*} |A|=\dfrac{1}{b-a}\left|\int_{a}^{b}f(x)dx\right|=\dfrac{1}{b-a}\int_{a}^{b}zf(x)dx=\dfrac{1}{b-a}\int_{a}^{b}u(x)dx+i\dfrac{1}{b-a}\int_{a}^{b}v(x)dx, \end{align*} so \begin{align*} \int_{a}^{b}v(x)dx=0, \end{align*} and hence \begin{align*} |A|\leq\dfrac{1}{b-a}\int_{a}^{b}u^{+}(x)dx\leq\dfrac{1}{b-a}\int_{a}^{b}|f(x)|dx=|A|, \end{align*} hence \begin{align*} \int_{a}^{b}(|f(x)|-u^{+}(x))dx=0, \end{align*} once again $|f(\cdot)|-u^{+}(\cdot)\geq 0$, so $|f|=u^{+}$. Then $u^{+}+u^{-}=|u|\leq|f|=u^{+}$ implies $u^{-}=0$. Also, $|f|^{2}=(u^{+})^{2}+|v|^{2}=|f|^{2}+|v|^{2}$ implies $v=0$, so $zf=u^{+}=|f|=|A|$.

Substitute $f=z^{-1}|A|$ into \begin{align*} A=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx, \end{align*} one gets $z^{-1}=\dfrac{A}{|A|}$, so $f=A$.

Answer 2

HINT: You can take the absolute value, multiply by $(b-a)$ and convert the left side to an integral: $$\int_a^b |A| dx = \left|\int_a^b f(x) dx\right|.$$