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Assume $\alpha(.)$ is a function defined on the bounded measurable set $E$.Let $$(Tx)(t)=\alpha(t)x(t)\ \ \ \ \ \ \ \ \ \ \ \ x\in L^2(E)$$.Then $T$ is a bounded linear operator from $L^2(E)$ to itself iff $\alpha(.)$ is measurable on $E$ and essential bounded.

How to prove this,thank you in advance.

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up vote 3 down vote accepted

That $T$ is linear is obvious: $T(cx + y) = \alpha(t)\bigl(cx(t) + y(t)\bigr) = c\alpha(t)x(t) + \alpha(t)y(t) = cTx + Ty$. Now, if $\alpha$ is essentially bounded, then there exists $M > 0$ such that the measure of the set of all $t$ where $|\alpha(t)|> M$ is zero. Therefore, outside a set of zero measure, $|Tx| = |\alpha(t)x(t)| = |\alpha(t)||x(t)|\leq M|x(t)|$. Therefore $Tx\in L^2(E)$. Finally, to prove that $T$ is bounded, by the above inequality, we have

$$\int_E|Tx|^2d\mu \leq M^2\int_E|x|^2 d\mu $$

In other words, $\|Tx\|_{L^2}\leq M\|x\|_{L^2}$. Hence, by definition, the operator $T$ is bounded, and $\|T\|\leq M$.

EDIT: The other direction.

Assume now that $T$, as defined, is a bounded linear operator on $L^2(E)$. Assuming that $\alpha$ is not essentially bounded on $E$, we get that for $\epsilon > 0$ there exists a positive measure set $\widetilde{E}\subset E$ such that $|\alpha(t)| > (\|T\| + \epsilon)^{1/2}$ for all $t\in \widetilde{E}$. Moreover, since $E$ is bounded, so is $\widetilde{E}$, so $0 < \mu(\widetilde{E}) < \infty$. Now take $x(t) = \mu(\widetilde{E})^{-1/2}\chi_{\widetilde{E}}\in L^2(E)$. We get:

$$\int_E |Tx|^2d\mu = \int_E|\alpha|^2|x|^2d\mu = \frac{1}{\mu(\widetilde{E})}\int_{\widetilde{E}}|\alpha|^2d\mu > \|T\| = \|T\|\|x\|_{L^2}.$$

This contradicts the fact that $\|T\|$ is the norm of $T$.

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Thank you WNY,how to prove the reverse?I will try it myself at the same time:)) – Jiangnan Yu Mar 22 '12 at 8:49
Oh, sorry, I missed that it was 'iff'. See the edit above. – William Mar 22 '12 at 9:32
PS. Actually, I neglected raising the integrals to $1/2$, but the adjustments should be obvious. – William Mar 22 '12 at 9:35
Ah thank you for your detailed answer.I learned it just now:) In order to raising the integrals to 1/2 we only need let $|\alpha(t)| > (\|T\|^2 + \epsilon)^{1/2} $ – Jiangnan Yu Mar 24 '12 at 7:36

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