A question on the homogeneous subspaces on the unit circle

Question

A homogeneous subspace on the unit circle $\mathbb{T}$ is a linear subspace $E$ of $L^1(\mathbb{T})$ having a norm $\|\cdot\|_E\geq\|\cdot\|_{L^1}$ under which it is a Banach space and satisfying in

1. $\|f_t\|_E=\|f\|_E$ where $f_t(x)=f(x-t)$.

2. $t\to f_t$ is continous.

Question. Let $E$ be a homogeneous subspace on $\mathbb{T}$. Then $\|F_n*f-f\|_E\to0$ where $$F_n(t)=\sum_{-n}^n\left(1-\frac{|k|}{n+1}\right)e^{ikt},$$ is the Fejer's kernel.

Answer 1

First, if $f\in E$ and $K\in L^1$ then $K*f$ is defined to be a vector-valued integral $$K*f=\frac1{2\pi}\int_{-\pi}^\pi K(t)f_t\,dt.$$

Lemma. $\|K*f\|_E\le\|K\|_1\|f\|_E$.

Proof: $\|K*f\|_E\le\frac1{2\pi}\int_{-\pi}^\pi|K(t)|\,\|f_t\|_E=\|f\|_E\frac1{2\pi}\int_{-\pi}^\pi |K(t)|\,dt$.

Now suppose $f\in E$ and $\epsilon>0$. Choose $\delta\in(0,\pi)$ so $|t|<\delta$ implies $\|f-f_t\|_E<\epsilon$. If $N$ is large enough then $F_N(t)<\epsilon$ for $\delta<|t|<\pi$; now $F_N*f-f=\frac1{2\pi}\int_{-\pi}^\pi F_N(t)(f_t-f)\,dt$, so $$\begin{align}\|F_N*f-f\|_E&\le\frac1{2\pi}\int_{-\delta}^\delta |F_N(t)|\|f_t-f\|_E +\frac1{2\pi}\int_{\delta<|t|<\pi} |F_N(t)|\|f_t-f\|_E \\&\le\epsilon\frac1{2\pi}\int_{-\delta}^\delta |F_N(t)|+2\|f\|_E\frac1{2\pi}\int_{\delta<|t|<\pi}|F_N(t)| \\&\le(1+2\|f\|_E)\epsilon.\end{align}$$

Aha: Above I say the convolution is defined as a vector-valued integral. Well I hope it was so defined, but what if it was defined as the almost-everywhere convergent pointwise integral? That makes various things harder. It took me a little while to show the two definitions are equivalent:

Suppose $f\in E$ and $K\in L^1$, and define $K*f$ as above. Then $K*f(x)=\frac1{2\pi}\int_{-\pi}^\pi K(t)f(x-t)\,dt$ for almost every $x$.

Trick proof: Note first that since $f\in L^1$ the integral converges for almost every $x$, and if we define $$g(x)=\frac1{2\pi}\int_{-\pi}^\pi K(t)f(x-t)\,dt$$then $g\in L^1$.

Fix $n\in\Bbb Z$ and define $\Lambda\in E^*$ by $$\Lambda h=\hat h(n)$$(note that $\|\cdot\|_E\ge\|\cdot\|_1$ shows that in fact $\Lambda\in E^*$). A change of variables shows that $$\Lambda f_t=e^{-int}\Lambda f;$$hence $$\Lambda(K*f)=\frac1{2\pi}\int_{-\pi}^\pi K(t)\Lambda(f_t) =\hat f(n)\hat K(n).$$So $K*f$ and $g$ have the same Fourier coefficients, hence $K*f=g$.