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

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 $$p<\infty$$.

• For any infinite measure space $L^{2} \cap L^{\infty}$ is not dense in $L^{\infty}$. – Kavi Rama Murthy Jun 23 at 5:16
• exists a counterexample? – eraldcoil Jun 23 at 5:17
• 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. – Kavi Rama Murthy Jun 23 at 5:20
• @KaviRamaMurthy Can you please check my answer? – 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 $$𝐶_𝑐 ⊂ \operatorname \cap_{ q = 1}^\infty 𝐿^𝑞⊂𝐿^2∩𝐿^𝑝$$ and $$𝐶_𝑐$$ is dense in $$L^p,\forall 1 \le p < \infty$$.

In case of $$𝑝=∞$$,

Note that $$f \in 𝐿^2\cap 𝐿^\infty \implies 𝑓∈𝐿^𝑝,∀2≤𝑝≤∞$$. 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 $$𝐿^𝑝→𝐿^∞,∀2≤𝑝<∞$$ but fail to be Bounded when $$𝐿^∞→𝐿^∞$$.

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 $$L^\infty$$