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.

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    $\begingroup$ 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.) $\endgroup$
    – user711689
    Apr 13, 2021 at 3:20
  • $\begingroup$ @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 $g\in L^\infty$. $\endgroup$
    – MaoWao
    Apr 13, 2021 at 14:13
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    $\begingroup$ You're right, I messed up the exponents. But same issue... $\endgroup$
    – user711689
    Apr 13, 2021 at 14:16
  • $\begingroup$ 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 $\mathbb{R}$. $\endgroup$
    – user711689
    Apr 13, 2021 at 14:23


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