Consider the space $L^\infty(\mathbb{R})$ and its (closed) subspace $S$ of $2 \pi$ periodic functions.
Define $\lVert f \rVert_\infty = \text{ess sup}_{x \in \mathbb{R}} |f(x)|$.
Now consider the quotient space $Q=L^\infty(\mathbb{R})/S$. Define the norm of the equivalence class $[f] = \left\{ g \in L^\infty: g-f \in S \right\}$ as $\lVert [f] \rVert_{Q} = \inf \left\{ \lVert f + g \rVert_\infty: g \in S \right\} = \text{dist}\left(f, S \right)$.
Calculate $\lVert \left[ \sin\frac{x}{2} \right] \rVert_Q $.
I know that $\lVert [f] \rVert_Q \leq \lVert f \rVert_\infty $. So $\lVert \left[ \sin\frac{x}{2} \right] \rVert_Q \leq 1$. The $Q$ norm can be expressed in several ways but taking the infimum is where I struggle. How do I proceed from here? Thanks.
Best Answer
I would guess that the best we can do is the $0$ function, simply because the graph of $\sin(x/2)$ oscillates negative and positive orientations of the same curve over the period $2\pi$. That is, $f(x) = \sin(x/2)$ satisfies $f(x + 2\pi) = -f(x)$. If we wish to uniformly approximate $f$ by a $2\pi$ period function, any attempt to fit the function to $f$ better than just the $0$ function over $[0, 2\pi]$ will make for a worse approximation over $[2\pi, 4\pi]$, and vice-versa.
Let's prove this intuition correct. Let's suppose that we can find a function $g \in S$ such that $\|f - g\|_\infty < 1$. Let $$\alpha = \frac{1 + \|f - g\|_\infty}{2} \in (0, 1).$$ Then, there is a non-trivial interval $I \subseteq [0, 2\pi]$ on which $f(x) > \alpha$ (simply because $\alpha < 1$, $f$ is continuous, and $f$ achieves a maximum of $1$). We then have, \begin{align*} \|f - g\|_\infty &= \operatorname{ess sup}_{x \in \Bbb{R}} |f(x) - g(x)| \\ &\ge \operatorname{ess sup}_{x \in \Bbb{R}} (f(x) - g(x)) \\ &\ge \operatorname{ess sup}_{x \in I} (f(x) - g(x)) \\ &\ge \operatorname{ess sup}_{x \in I} (\alpha - g(x)) \\ &= \frac{1 + \|f - g\|_\infty}{2} - \operatorname{ess inf}_{x \in I} g(x), \end{align*} hence $$\operatorname{ess inf}_{x \in I} g(x) \ge \frac{1 - \|f - g\|_\infty}{2} > 0.$$ On the other hand, note that $I + 2\pi$ is another non-trivial interval, and $$x \in I + 2\pi \implies f(x) < -\alpha.$$ Then \begin{align*} \|f - g\|_\infty &\ge \operatorname{ess sup}_{x \in \Bbb{R}} (g(x) - f(x)) \\ &\ge \operatorname{ess sup}_{x \in I + 2\pi} (g(x) - f(x)) \\ &\ge \operatorname{ess sup}_{x \in I + 2\pi} (g(x) + \alpha) \\ &= \operatorname{ess sup}_{x \in I + 2\pi} g(x) + \frac{1 + \|f - g\|_\infty}{2}, \end{align*} hence $$\operatorname{ess sup}_{x \in I + 2\pi} g(x) \le -\frac{1 - \|f - g\|_\infty}{2} < 0.$$ But, because $g$ is periodic with period $2\pi$, we must have $$0 < \operatorname{ess inf}_{x \in I} g(x) \le \operatorname{ess sup}_{x \in I} g(x) = \operatorname{ess sup}_{x \in I + 2\pi} g(x) < 0,$$ which is a contradiction. Therefore, the norm of the equivalence class of $f$ is indeed $1$.