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I am reading a paper which deals with the densities of $(X_t)$ where $X_t$ is typically a Levy process.

The paper assumes that the density of, say, $X_1$, is square-integrable, i.e. in $L^2$. Other times, it will assume that the characteristic function of $X_1$ is integrable, i.e. in $L^1$.

It provides no reference to these facts.

Is this known to be true? Is this in general true for densities that one typically encounters? Or are these results particularly true for Levy processes? If so, does it hold for all of them, or must some conditions be satisfied?

saei
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  • Note that if the characteristic function is in L^1 then the density automatically exists and is in L^2. That’s because the characteristic function is bounded, so also being in L^1 automatically implies being in L^2 by interpolation. Then one uses the fact that the Fourier transform is an isometry on L^2. Of course not every Levy process has a characteristic function in L^1, e.g. Poisson process. However, if the Brownian component of the Levy process is nonzero, then both assumptions will hold true. – shalop Feb 26 '20 at 20:23

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Not all densities are square-integrable, and not all characteristic functions are integrable. Counterexamples are not hard to construct. The fact that the paper is stating an assumption rather than a fact should tell you something.

Robert Israel
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    How about for Levy processes specifically? I know we have the Levy Khinchinte theorem which narrows down the characteristic function quite a lot. Do conditions exist for integrability of this characteristic function? – saei Feb 26 '20 at 14:00