Continuous extension of a bounded linear operator in Fourier multiplier definition.

Question

In $\mathbb{R}^n$. For $1\leq p<\infty$ with $T:L^2\cap L^p\to L^p$ bounded linear operator, bounded in $L^p$. I know that it is possible to extend $T$ to $\bar{T}$ such that $T:L^p\to L^p$, bounded in $L^p$. This, by Bounded Linear Theorem.

My doubts are: Is $L^p$ the completion of $L^2\cap L^p$? ($1\leq p<\infty)$

With $p=\infty$. Is it possible to extend the operator $T$? Or this fails?

I ask this because in Stein's 'Singular integral and differentiability properties', in fourier multiplier definition, a operator $T_m:L^2\cap L^p\to L^p$ is extended to $T_m:L^p\to L^p$ when $p<\infty$.

Answer 1

For $1 \le <∞$,

$T : L^2 \cap L^p \to L^p$ bounded $\implies$ $T$ extends to a bounded linear operator $L^p \to L^p$ as $$_ ⊂ \operatorname \cap_{ q = 1}^\infty ^⊂^2∩^$$ and $_$ is dense in $L^p,\forall 1 \le p < \infty$.

In case of $=∞$,

Note that $f \in ^2\cap ^\infty \implies ∈^,∀2≤≤∞$. Hence in order to find a bounded linear transformation $T : L^2 \cap L^\infty \to L^\infty$ that fails to extend as a bounded operator on $L^\infty$, combining with previous part, one basically has to produce a linear transformation that will extend to a bounded linear transformation from $^→^∞,∀2≤<∞$ but fail to be Bounded when $^∞→^∞$.

What can be more natural than convolution operators? For any $f \in L^p(\Bbb R^n), g \in L^q(\Bbb R^n) $ we have from Young's inequality $$||f*g||_r \le ||f||_p ||g||_q \text{ where } \frac{1}{p}+\frac{1}{q}=1+\frac{1}{r} \text{ and } 1\le p,q \le r \le \infty$$

Comment: Moreover, for a fixed $q,r$ in the above relations, you can show by doing a bit of Dimensional analysis ( i.e. dilating the functions ) that $p$ is indeed the unique choice.

So just choose a function $g \in L^q(\Bbb R^n), \forall 1 < q \le 2$ but $g \notin L^1(\Bbb R^n)$ say $g(x)=\frac{1}{||x||} \Bbb{1}_{\{x \in \Bbb R^n: ||x||>1\}}$

Now define the transformation $T_g : L^2 \cap L^\infty \to L^\infty$ by $T_g(f):= f*g $, then by Young's inequality and the adjacent comment, it follows that $T_g$ is indeed a linear transformation with the desired properties i.e. Bounded from $L^2 \cap L^\infty \to L^\infty$ but does not extend to a Bounded operator on $L^\infty$