On translation invariant subspace of $L^1(\mathbb T)$

Question

A subspace $M \subset L^p(\mathbb{T})$ is called translation invariant if $f \in M$ implies that $\tau_\theta f \in M$ for all $\theta \in [-\pi, \pi]$.

1. Suppose that $M$ is a closed translation invariant subspace of $L^1(\mathbb{T})$. If $f \in L^1(\mathbb{T})$, then prove that $f*g \in M$ for all $g \in M$, i.e. $M$ is an ideal.
2. Suppose that $M$ is a closed subspace of $L^1(\mathbb{T})$ with the property that for all $f \in L^1(\mathbb{T})$ and $g \in M$, $f * g \in M$. Prove that $M$ is translation invariant.
3. Do these results hold for $L^p(\mathbb{T})$, $1 < p < \infty$?

Here, $\tau_\theta f(x) = f(x - \theta)$ and $\mathbb T$ is the unit circle.

I have used some of the given hints to notice that if $M$ is a closed subspace of a Banach space, then $v ∈ M$ if $ψ ∈ X^∗$ and $ψ(M) = 0$ implies $ψ(v) = 0$. Also, I have proved that for $f,g\in L^1(\mathbb T)$ and $\theta\in[-\pi,\pi]$, we have $\tau_{\theta_0}(f\ast g)=\left(\tau_{\theta_0}(f)\right)\ast g=f\ast \left(\tau_{\theta_0}(g)\right)$ (which is supposed to help in part (b)).

But, I don't have any ideas to continue.

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Parts (a) and (b) has been solved while I still have doubts with (c). Please see if that can be solved as well.

Answer 1

For proving (2), find an approximate identity $\{K_n\}$, so that $f * K_n \to f$ in $L^1$ for all $f \in L^1$. Now, given $g \in M$ and some $\theta$, we have $(\tau_\theta g) * K_n \to \tau_\theta g$ in $L^1$. But each term in the sequence can be rewritten as $g * (\tau_\theta K_n)$, which is in $M$ by hypothesis. Since $M$ is closed in $L^1$, we have $\tau_\theta g \in M$.

For proving (1) using the hint, suppose that $\psi \in (L^1(\mathbb{T}))^*$ and $\psi(M) = 0$. Then, the Riesz Representation Theorem gives a unique Radon measure $\mu$ such that $$ \psi(f) = \int_{\mathbb{T}} f\:d\mu$$ for all $f \in L^1$.

Now for $g \in M$ and $f \in L^1$, calculate $$\psi(f * g) = \iint f(t) g(\theta - t) \:dt\:d\mu(\theta).$$ Invoking Fubini-Tonelli, this is $$\int\left(\int g(\theta - t)\:d\mu(\theta)\right) f(t)\:dt = \int \psi(\tau_t g) f(t)\:dt.$$ But $\tau_t g \in M$, hence $\psi(\tau_t g) = 0$, so $\psi(f * g) = 0$ as desired.

Answer 2

The answer to $1$ for $1\le p<\infty$ can be as follows. For $g\in L^p$ the function $t\mapsto \tau_tg$ is continuous with respect to $L^p$-norm. Let $t_k={\pi k\over n}.$ Assume $f\in C(\mathbb{T}).$ We have $$S_n:= {1\over 2n}\sum_{k=-n+1}^{n}f(t_k)\tau_{t_k}g\in M$$ Let $$A_n=\sup_{|t-s|\le \pi /n} |f(t)-f(s)|\quad B_n=\sup_{|t-s|\le \pi /n} \|\tau_tg-\tau_sg\|_p$$ Then $$f*g-S_n={1\over 2\pi}\sum_{k=-n+1}^n\,\,\int\limits_{t_{k-1}}^{t_k} [f(t)\tau_tg-f(t_k)\tau_{t_k}g]\,dt\\ = {1\over 2\pi}\sum_{k=-n+1}^n\,\,\int\limits_{t_{k-1}}^{t_k} [f(t)-f(t_k)]\tau_tg\,dt+{1\over 2\pi}\sum_{k=-n+1}^n\,\,\int\limits_{t_{k-1}}^{t_k} f(t_k)[\tau_tg-\tau_{t_k}g]\,dt$$ Thus $$\|f*g-S_n\|_p\le A_n\|g\|_p+B_n\|f\|_\infty\to 0$$ Hence $f*g\in M.$

If $f\in L^1$ then there exists $f_n\in C(\mathbb{T})$ such that $\|f-f_n\|_1\le {1\over n}.$ Hence $$\|f*g-f_n*g\|_p\le \|f-f_n\|_1\|g\|_p\le {1\over n}\|g\|_p\to 0$$ We know that $f_n*g\in M.$ Thus $f*g\in M$ as $M$ is closed.

The proof of $2$ can be performed by approximating pointwise the $\delta_0$ function by a sequence of continuous nonnegative functions $f_n$ with integral equal $1.$ In this way the sequence of measures $f_n(t)\,dt$ tends weakly to the probability measure concentrated at $0.$