Let $S$ be an $L$ shaped region in the unit cube $Q:=[0,1]\times [0,1]$:
$$
S:=Q\backslash C,\quad C:=\left[\frac 1 2,1\right]\times \left[\frac 1 2,1\right].
$$
Consider the multiplier operator $T$ defined by
$$
\widehat {Tf}(\xi):=\hat f(\xi)1_{S}(\xi)
$$
Then what can we say about the operator norm of $T$ on $L^p(\mathbb R^2)$, where $1<p<\infty$?

We have the following observations:

 1. If $p=2$, then the operator norm is 1, which follows trivially by the Plancherel theorem. Hence we consider $p\neq 2$ only.
 2. It is well known that any rectangle multiplier in $\mathbb R^2$ (by symmetries and the Hilbert transform) has operator norm $C_p$  which depends on $p$ only. 
 3. Therefore, since $L$ can be partitioned into two disjoint rectangles (in any way), it is easy to see that $T$ has operator norm bounded by $2$.
 4. Cordoba showed that for a polygon with $N$ sides, the polygon multiplier has norm bounded by $C(\log N)^{\alpha(p)}$. Moreover, this bound is sharp.

In the case of an $L$-shaped region, is it true that $\left\|T\right\|_p>1$?



<cite authors="Cordoba, A.">_Cordoba, A._, [**The multiplier problem for the polygon**](http://dx.doi.org/10.2307/1970926), Ann. Math. (2) 105, 581-588 (1977). [ZBL0361.42005](https://zbmath.org/?q=an:0361.42005).</cite> showed that for any polygon