Continuity of log(x) as a multiplication operator

Define a multiplication operator $$T_{log}f(x) = f(x)\cdot log(x)$$. For $$1 \le p < \infty$$ and the usual Lebesgue space $$L^p(\mathbb{R})$$, is $$T_{log}$$ a continuous operator from $$L^p(\mathbb{R})$$ to itself? I know this is not true for $$p = \infty$$ and I feel this should be true for finite $$p$$. But I don't know how to prove it or if there exist counter examples.

• In general, if $1 \leq p \leq \infty$ and $\frac{1}{p} + \frac{1}{q} = 1$, then the only way to get $\|fg\|_{L^{p}(X)} \leq C \|f\|_{L^{p}(X)}$ for all $f \in L^{p}(X)$ is if $\|g\|_{L^{q}(X)} \leq C$. (These are, in fact, equivalent statements.)
– user711689
Apr 13, 2021 at 3:20
• @PeterMorfe That's not correct. The multiplication operator is bounded from $L^p$ to $L^1$ if and only if $g\in L^q$, but it is bounded from $L^p$ to $L^p$ if and only if $g\in L^\infty$. Apr 13, 2021 at 14:13
• You're right, I messed up the exponents. But same issue...
– user711689
Apr 13, 2021 at 14:16
• The correct statement is, for $p,q,r \in [1,\infty)$ with $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$, the function $g$ has $\|fg\|_{L^{r}(X)} \leq C\|f\|_{L^{p}(X)}$ for all $f \in L^{p}(X)$ if and only if $g \in L^{q}(X)$ and $\|g\|_{L^{q}(X)} \leq C$. (Exercise using Holder's inequality and $L^{p}-L^{q}$ duality.) Above $r = p$ so $q = \infty$. At any rate, the logarithm isn't in any $L^{p}$ space on $\mathbb{R}$.
– user711689
Apr 13, 2021 at 14:23