Show that $x^2 + 3\sin^2 x$ satisfies the Polyak–Łojasiewicz condition

Question

A function $f \colon \mathbb{R}^d \to \mathbb{R}$ is said to satisfy the Polyak–Łojasiewicz condition (with constant $\mu >0$) if $$f(x) - \min\limits_{x \in \mathbb R} f(x) \leq \frac{1}{2\,\mu}\,\|\nabla f(x)\|^2 \quad \text{for all $x \in \mathbb{R}^d$}. $$ I have seen many literatures in machine learning saying/commenting that the function $f(x) = x^2 + 3\sin^2 x$ satisfies the Polyak–Łojasiewicz condition with $\mu = \frac{1}{32}$, while no detailed computations are given. I tried to establish the desired inequality but I didn't succeed. Can anyone provide a rigorous proof of the desired result?

Answer 1

If $f(x) = x^2 + 3\sin^2 x$, then ${\rm min}_{x\in{\mathbb R}}f(x) = f(0) = 0$ and $\|\nabla f(x)\|^2 = (2x+3\sin 2x)^2$. Therefore, proving the Polyak–Łojasiewicz condition with the condition $\mu=\frac{1}{32}$ is equivalent to proving that $g(x)\ge 0$ for all $x\in{\mathbb R}$, where $$ g(x) = 16\,(2x+3\sin 2x)^2 - (x^2 + 3\sin^2 x).\tag{1} $$ Since $\sin^2 x \leq x^2$ for all $x\in{\mathbb R}$, we have $$ g(x) \ge 16\,(2x+3\sin 2x)^2 - 4x^2 = 64x^2(1+ 3\,{\rm sinc}\,2x)^2 - 4x^2, \tag{2} $$ where ${\rm sinc}\,z = \frac{\sin z}{z}$. Let's now examine two cases:

1. If $|x|\leq\frac{\pi}{2}$, then ${\rm sinc}\, 2x \ge 0$ and $g(x)\geq 64x^2-4x^2\ge 0$;
2. If $|x|>\frac{\pi}{2}$, we use the inequality$^{(*)}$ ${\rm sinc}\,2x>-\frac{1}{4}$ in $(2)$ to derive $$ g(x) > 64x^2\left(1-\frac{3}{4}\right)^2 - 4x^2 = 0. $$

This concludes the proof of inequality $(1)$.

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$^{(*)}$See, for instance, https://en.wikipedia.org/wiki/Sinc_function.