The triangle inequality for integrals $$\left| \int_a^b f(t) dt\right| \le \int_a^b |f(t)|dt$$ is well-known, where $f(t):\mathbb R\to \mathbb C$. However, I was wondering if there is a condition for the inequality to be strict, i.e., LHS<RHS.
Background
I encountered this problem while I was studying Fourier Analysis:
Let $f\in L^1(\mathbb R)$ be a bounded continuous function such that $f(x)>0$ for all $x\in \mathbb R$, denote by $F$ the Fourier transform of $f$, defined by $$F(y)=\int_{\mathbb R}f(x)e^{-2\pi iyx}dx$$ Prove that $|F(y)|<F(0)$ for every $y\neq 0$.
From the triangle equality, it is trivial that $$|F(y)|=\left|\int_{\mathbb R}f(x)e^{-2\pi iyx}dx\right|\leq \int_{\mathbb R}|f(x)e^{-2\pi iyx}|dx=\int_{\mathbb R}f(x)dx=F(0)$$ But how can I prove that the inequality is strict?
My attempt
I'm not very good at real analysis (only have some knowledge of measure theory), but had a go like this:
Consider $y\neq 0$. Let $g(x,y)=f(x)e^{-2\pi iyx}$. We have $|g(x,y)|=f(x)$. The set $E_v=\{x\in \mathbb{R}|e^{-2\pi iyx}=1\}$ has measure zero. On the complement of $E_v$, we have $g(x,y)\neq |g(x,y)|$. I believe this would imply $$\left|\int_{\mathbb R \setminus E_v}g(x,y)dx\right|<\int_{\mathbb R \setminus E_v}|g(x,y)|dx$$ but couldn't prove it.
My thoughts
My instinct tells me that for $$\left| \int_a^b f(t) dt\right| = \int_a^b |f(t)|dt$$ to hold, $f(t)$ must be a positive multiple of a (real or complex) number almost everywhere, am I correct?
Let $\int_a^{b} f(t)dt=re^{i\theta}$ with $r \geq 0, \theta$ real. Then we have $\int_a^{b} e^{-i\theta}f(t)dt=r=\int_a^{b} |f(t)|dt$. Taking real parts we get
$\int_a^{b} \Re [e^{-i\theta}f(t)]dt=r=\int_a^{b} |f(t)|dt$. But $\Re [e^{-i\theta}f(t)] \leq |f(t)|$ so we must have $\Re [e^{-i\theta}f(t)] =|f(t)|$ almost evrywhere. This implies that $\Im [e^{-i\theta}f(t)] =0$ almost evrywhere and $e^{-i\theta}f(t) =|f(t)|$ almost everywhere.