Absolute Value of Complex Integral

Let $[a,b]$ be a closed real interval. Let $f:[a,b] \to \mathbb{C}$ be a continuous complex-valued function. Then $$\bigg|\int_{a}^{b} f(t)dt \ \bigg| \leq \int_{a}^{b} \bigg|f(t)\bigg| dt,$$ where the first integral is a complex integral, and the second integral is a definite real integral.

There's a neat "rotational" proof of this in D'Angelo's An Introduction to Complex Analysis and Geometry.

Question: Can this fact also be proven using the Cauchy-Schwarz Inequality? If so, some help would be nice.

Thank you...

Solution 1:

We first state the Cauchy-Schwarz inequality for definite integrals: Let $u$ and $v$ be real functions which are continuous on the closed interval $[a,b].$ Then $$\left(\int_{a}^{b} u(t)v(t)dt \right) \leq \left(\int_{a}^{b} u(t)dt \right)^2 \cdot \left(\int_{a}^{b} v(t)dt \right)^2$$ Since $F$ is complex-valued, we can write $F(t) = u(t) + iv(t),$ with $u(t), v(t)$ real-valued. Thus $|F(t)|^2 = u^2+v^2.$ Also \begin{align}\left(\int_{a}^{b} F(t)dt \right)^2 = \left(\int_{a}^{b} u(t) + iv(t) dt \right)^2. \end{align} Observe that \begin{align*} \left(\int_{a}^{b} u ~dt \right)^2 &= \left(\int_{a}^{b} \dfrac{u}{(u^2+v^2)^{1/4}}(u^2+v^2)^{1/4}dt \right)^2 \\ &\overset{\mathrm{C-S \ ineq.}}{\leq} \left(\int_{a}^{b} \dfrac{u}{(u^2+v^2)^{1/4}}dt \right)^2\left(\int_{a}^{b} \sqrt[4]{u^2+v^2} ~dt \right)^2 \\ &= \left(\int_{a}^{b} \dfrac{u^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right) \end{align*} and \begin{align*} \left(\int_{a}^{b} v ~dt \right)^2 &= \left(\int_{a}^{b} \dfrac{v}{(u^2+v^2)^{1/4}}(u^2+v^2)^{1/4}dt \right)^2 \\ &\overset{\mathrm{C-S \ ineq.}}{\leq} \left(\int_{a}^{b} \dfrac{v}{(u^2+v^2)^{1/4}}dt \right)^2\left(\int_{a}^{b} \sqrt[4]{u^2+v^2} ~dt \right)^2 \\ &= \left(\int_{a}^{b} \dfrac{v^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right). \end{align*} So \begin{align*} \bigg|\int_{a}^{b} F(t) ~dt \bigg|^2 &= \bigg|\int_{a}^{b} u(t) + iv(t) ~dt \bigg|^2 \\ &= \bigg|\int_{a}^{b} u(t) ~dt + i\int_{a}^{b}v(t) ~dt \bigg|^2 \\ &= \left(\int_{a}^{b} u ~dt \right)^2 + \left(\int_{a}^{b} v ~dt \right)^2 \\ &\leq \left(\int_{a}^{b} \dfrac{u^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right) \\ &+ \left(\int_{a}^{b} \dfrac{v^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right) \\ &= \left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right)\left(\int_{a}^{b} \dfrac{u^2+v^2}{\sqrt{u^2+v^2}}dt \right) \\ &= \left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right)^2 \\ &= \left(\int_{a}^{b} \bigg|F(t)\bigg|\right)^2 dt \end{align*} Now, recalling that the modulus is always non-negative we can take the square root of both sides and we arrive at the result.