About a question on measurable functions

Question

I came across this preliminary question in a textbook on Hardy spaces Let $f$ be a measurable function on $\mathbb{T}=\{z\in\mathbb{C}: |z|=1\}$, and let $$f_r(e^{it})=f(e^{i(t-r)}).$$ Show that $$\lim_{r\longrightarrow 0}\|f_r-f\|_X=0$$ if $X=L^p(\mathbb{T}), 1\leq p<\infty,$ or $X=C^n({\mathbb{T}})$. Show that this property does not hold if $X=L^\infty(\mathbb{T})$ or $X=Lip_{\alpha}(\mathbb{T})$, $0<\alpha\leq 1.$

Answer 1

The continuous functions $C(\mathbb{T})$ are dense in $L^p$ for $1 \le p < \infty$, but not for $p=\infty$.

Let $1 \le p < \infty$ be fixed, and let $f \in L^p(\mathbb{T})$. Let $\epsilon > 0$ be given. Then there exists $g \in C(\mathbb{T})$ such that $\|f-g\|_p < \frac{\epsilon}{3}$, which also gives $\|f_r - g_r \|_p < \frac{\epsilon}{3}$ for all $r$. Because $g$ is uniformly continuous on $\mathbb{T}$, then there exists $r_0 \in (0,1)$ such that $\|g-g_r\|_p < \frac{\epsilon}{3}$ for all $0 \le r < r_0$. Hence, $$ \|f-f_r\|_p \le \|f-g\|_p + \|g-g_r\|_p+\|g_r-f_r\|_p < \epsilon,\;\; \mbox{ whenever } r_0 \le r < 1. $$ That proves that $\lim_{r\rightarrow 0}\|f-f_r\|_p=0$ for $1 \le p < \infty$.

To show that the result does not hold for $L^{\infty}(\mathbb{T})$, let $f$ be $1$ on half of $\mathbb{T}$ and $0$ on the other half. Then $\|f-f_r\|_{\infty}=1$ for all $0 < r < 2\pi$.