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[ Remark: The following question was asked yesterday, and obtained 3 votes. Unfortunately it has been deleted by the OP overnight without receiving any answers.]

Let $f:\mathbb C \to \mathbb R$ be a continuous real-valued function. Suppose for all $z \in \mathbb C$, we have $|f(z)|\leq1$. Show that

$$\left|\int_C f(z)\, dz\right|\leq4$$ where $C$ is the unit circle traversed counterclockwise.

I used this relationship but the smallest I can get is $2\pi$: $$\left|\int f(z)\, dz\right| \leq \int |f(z)|\, |dz| \leq \sup |f(z)|\cdot L = 2\pi\sup |f(z)| = 2\pi$$

Christian Blatter
  • 226,825

1 Answers1

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The function $g(t):=f\bigl(e^{it}\bigr)\in[{-1},1]$ is $2\pi$-periodic. By definition of the line integral the parametrization $t\mapsto z(t):=e^{it}$ $(0\leq t\leq2\pi)$ gives $$\int_C f(z)\>dz=\int_0^{2\pi} g(t)\>ie^{it}\>dt=:i\rho \,e^{i\alpha}$$ for some $\rho\geq0$ and $\alpha\in{\mathbb R}$. We have to prove that $\rho\leq4$. To this end consider $$\int_0^{2\pi}g(t+\alpha)e^{it}\>dt=\int_0^{2\pi}g(\tau)e^{i(\tau-\alpha)}\>d\tau=\rho\ .$$ Since $g$ is real-valued we can conclude that in fact $$\rho=\int_0^{2\pi} g(t+\alpha)\,\cos t\>dt\leq\int_0^{2\pi}\bigl|\cos t\bigr|\>dt=4\ .$$

Christian Blatter
  • 226,825
  • Very nice answer, some really basic questions: did you say that $g$ is $2\pi$-periodic because it is function only of $e^{it}$ and $e^{it}$ is a $2\pi$-periodic function? You say that the integral is equal to $i \rho e^{i\alpha}$ because $\rho e^{i\alpha}$ is a generic complex number so, when you say "for some $\rho \geq 0$ and $\alpha \in \mathbb{R}$" you mean "there exists $\rho \geq 0$ and $\alpha \in \mathbb{R}$ such that the integral is in the form of $\rho e^it$? Is $g$ real-valued because $f:\mathbb{C} \to \mathbb{R}$? Where the modulus in the last inequality comes from? Thanks. – ZaWarudo Apr 26 '20 at 02:07
  • @ZaWarudo: Except the last question: Everything Yes. Last question: This is a standard move in the estimation of integrals. Note that $|g(t)|\leq1$. – Christian Blatter Apr 26 '20 at 09:07
  • thanks again for your kindness, I actually understood that we used $\left| \int_0^{2\pi} g(t+\alpha) \cos t \text{d}t \right| \leq \int_0^{2\pi} |g(t+\alpha) \cos t| \text{d}t \leq \int_0^{2\pi} |\cos t| \text{d}t$. If this is correct, I only have a doubt on why we put the modulus on, no doubts about the estimation. The only thing that comes into my mind is that $\rho \geq 0$, so when we put $\rho=\int_0^{2\pi} g(t+\alpha) \cos t \text{d}t$ we have to impose that the integral is positive and so we consider its modulus, am I right or am I wrong? – ZaWarudo Apr 26 '20 at 18:01
  • When you put $\tau :=t+\alpha$, why did the limits of the integral not change? – uno Mar 24 '21 at 06:49
  • @uno: Because $g$ is $2\pi$-periodic. – Christian Blatter Mar 24 '21 at 07:59