When I can safely assume that a function is a Laplace transform of other function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:37:45Z http://mathoverflow.net/feeds/question/44713 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-function When I can safely assume that a function is a Laplace transform of other function? Rorsa 2010-11-03T19:24:39Z 2010-12-03T07:55:11Z <p>If I have a function and I want to represent it as being the Laplace transform of another, that is, I want to be sure that there is $\hat{f}(s)$ such that my function $f(x)$ can be written as:</p> <p>$f(x) = \int ds \hat{f}(s) \exp(-sx)$</p> <p>what conditions should I impose over $f(x)$? </p> <p>In other words, what are the conditions for the Fourier-Mellin-Bromwich integral</p> <p>$\hat{f}(s) = \frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} f(x) \exp(sx) dx$</p> <p>to exist?</p> http://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-function/44972#44972 Answer by Stopple for When I can safely assume that a function is a Laplace transform of other function? Stopple 2010-11-05T17:41:52Z 2010-11-05T17:41:52Z <p>In Audrey Terras' "Harmonic Analysis on Symmetric Spaces and Applications, I" she has on p. 21 the following: Suppose $\exp(-cx)f(x)$ lies in $L^1(\mathbb R)$ and $f(x)$ vanishes for negative $x$. Assume also that $f(x)$ is piecewise differentiable. Then </p> <p>$(f(x+)+f(x-))/2=\lim_{T\to\infty}\frac{1}{2\pi i}\int_{c-i T}^{c+i T}\exp(sx)\mathcal Lf(s)ds$</p> <p>(NB traditional Laplace transform notation seems to be the reverse of yours; $x>0$ and $s\in\mathbb C$)</p> http://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-function/44988#44988 Answer by Andrey Rekalo for When I can safely assume that a function is a Laplace transform of other function? Andrey Rekalo 2010-11-05T19:35:45Z 2010-11-06T12:20:48Z <p>The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1)$$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau&lt;\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended by $0$ to $t\leq 0$) belonging to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)\cap L^2(\mathbb R)$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t&lt;0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), &amp; t>0 \\ \\ 0, &amp; t\leq 0 \end{cases}$$</p> <p>One can show also that the Parseval identity holds</p> <p>$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt,$$ so there is a complete analogy with the standard Fourier transform.</p> <hr> <p><strong>Executive summary.</strong> A function $f$ is the Laplace transform of some function $\phi$ satisfying condition (1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the <a href="http://en.wikipedia.org/wiki/Hardy_space" rel="nofollow">Hardy space</a> on a (right) half-plane. </p> http://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-function/48149#48149 Answer by Zen Harper for When I can safely assume that a function is a Laplace transform of other function? Zen Harper 2010-12-03T07:53:19Z 2010-12-03T07:53:19Z <p>This kind of question is very interesting, and I too would like to know answers.</p> <p>Sorry to self-publicise; I hope it's not regarded as impolite, but since I have also considered this exact kind of question, it's quickest just to refer to my own paper (<em>and the references I give in there</em>):</p> <p><strong>Laplace Transform Representations and Paley-Wiener Theorems for Functions on Vertical Strips</strong> by Zen Harper. <em>Documenta Math. 15 (2010) 235-254</em>.</p> <p>This is <em>freely available online</em> from the journal:</p> <p><a href="http://www.math.uiuc.edu/documenta/vol-15/vol-15-eng.html" rel="nofollow">http://www.math.uiuc.edu/documenta/vol-15/vol-15-eng.html</a></p> <p>Given an analytic function on a vertical strip, I try to find conditions which guarantee it can be represented by a bilateral Laplace transform (in various senses). I definitely don't claim to have any kind of complete answer, but it's the best I know of (<em>by definition! If I knew any better answers, I would have written them in my paper!</em>)</p> <p>It seems like there are still many open questions about this.</p>