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 $p<\infty$.

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    $\begingroup$ For any infinite measure space $L^{2} \cap L^{\infty}$ is not dense in $L^{\infty}$. $\endgroup$ – Kavi Rama Murthy Jun 23 at 5:16
  • $\begingroup$ exists a counterexample? $\endgroup$ – eraldcoil Jun 23 at 5:17
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    $\begingroup$ On the real line you cannot approximate the constant function $1$ by an $L^{2}$ function w.r.t. $L{^\infty}$ norm because $\|1-g\|_{\infty} <\frac 1 2 $ implies $|g(x)| \geq \frac 1 2$ almost evreywhere. $\endgroup$ – Kavi Rama Murthy Jun 23 at 5:20
  • $\begingroup$ @KaviRamaMurthy Can you please check my answer? $\endgroup$ – Brozovic Aug 1 at 10:41

For $1 \le 𝑝 <∞$,

$T : L^2 \cap L^p \to L^p$ bounded $\implies$ $T$ extends to a bounded linear operator $L^p \to L^p$ as $$𝐢_𝑐 βŠ‚ \operatorname \cap_{ q = 1}^\infty 𝐿^π‘žβŠ‚πΏ^2∩𝐿^𝑝$$ and $𝐢_𝑐$ is dense in $L^p,\forall 1 \le p < \infty$.

In case of $𝑝=∞$,

Note that $f \in 𝐿^2\cap 𝐿^\infty \implies π‘“βˆˆπΏ^𝑝,βˆ€2β‰€π‘β‰€βˆž$. 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 $𝐿^𝑝→𝐿^∞,βˆ€2≀𝑝<∞$ but fail to be Bounded when $𝐿^βˆžβ†’πΏ^∞$.

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 $L^\infty$

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