User:Tkojar/Sandbox/Lerch's Theorem

Statement
Suppose \(f\in{}L^{p}((0,\infty),e^{-at})\) for \(1\le{}p\le\infty\) and \(\mathcal{L}\{f\}(s)=0\) for all \(s>a\). Then \(f=0\) almost everywhere on \((0,\infty)\).

Proof
We prove this theorem in steps. First we demonstrate this for \(f\in{}C_{0}((0,\infty))\). Then we use the previous case to demonstrate the theorem for \(f\in{}L^{p}((0,\infty))\) for \(1\le{}p\le\infty\). Finally, we use the previous step to demonstrate this for \(f\in{}L^{p}((0,\infty),e^{-at})\).

Step \(1\): Let \(f\in{}C_{0}((0,\infty))\) and suppose that \(\mathcal{L}\{f\}(s)\) for all \(s>0\) (note that the Laplace transform is defined since \(f\in{}L^{p}((0,\infty))\) for all \(1\le{}p\le\infty\)).Observe that \begin{equation*} 0=\mathcal{L}\{f\}(s)=\int_{0}^{\infty}\!{}e^{-st}f(t)\mathrm{d}t =-\int_{0}^{\infty}\!{}\bigl(e^{-t}\bigr)^{s-1} f\bigl(-\ln\bigl(e^{-t}\bigr)\bigr)(-1)e^{-t}\mathrm{d}t \end{equation*} and so making the substitution \(u=e^{-t}\) so that \(\mathrm{d}u=-e^{-t}\mathrm{d}t\) gives \begin{equation*} 0=\int_{0}^{1}\!{}u^{s-1}f\bigl(-\ln(u)\bigr)\mathrm{d}u. \end{equation*} Observe that since \(f\in{}C_{0}((0,\infty))\) then \(g(u)=f\bigl(-\ln(u)\bigr)\) extends to a continuous function defined on \([0,1]\) by defining \(g(0)=0=g(1)\). In particular, we have by choosing \(s=1,2,3,4,\ldots\) that \begin{equation*} 0=\int_{0}^{1}\!{}u^{n}g(u)\mathrm{d}u,\,\,\,n=0,1,2,\ldots, \end{equation*} where the integral is understood as over the compact interval \([0,1]\). By the Weierstrass Approximation Theorem we obtain that \(g\equiv0\). Thus, since \(-\ln(u)\) is a bijection between \((0,1)\) and \((0,\infty)\), we have that \(f\equiv0\). Step \(2\): Now suppose \(f\in{}L^{p}((0,\infty))\) for \(1\le{}p\le\infty\). We extend \(f\) to a function \(\tilde{f}:\mathbb{R}\to\mathbb{R}\) by defining \begin{equation*} \tilde{f}(t)= \begin{cases} f(t)& t>0\\ & \\ 0& t\le0 \end{cases} \end{equation*}

Now we define \(\tilde{f}_{\epsilon}:(0,\infty)\to\mathbb{R}\), for \(\epsilon>0\), by \begin{equation*} \tilde{f}_{\epsilon}(t)=\int_{0}^{\infty}\!{} \phi_{\epsilon}\bigl(t-y\bigr)\tilde{f}(y)\mathrm{d}y \end{equation*} where \(\phi_{\epsilon}(t)=\frac{1}{\epsilon}\phi\left(\frac{t}{\epsilon}\right)\) and \(\phi:\mathbb{R}\to\mathbb{R}\) is a non-negative, smooth, function supported in \([0,1]\) such that \(\int_{\mathbb{R}}\!{}\phi(t)\mathrm{d}t=1\). Observe that: \begin{align*} \mathcal{L}\{\tilde{f}_{\epsilon}\}(s)&= \int_{0}^{\infty}e^{-st}\tilde{f}_{\epsilon}(t)\mathrm{d}t= \int_{0}^{\infty}e^{-st} \int_{0}^{\infty}\phi_{\epsilon}(t-y)\tilde{f}(y)\mathrm{d}y\mathrm{d}t\\ &=\int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)\tilde{f}(y)\mathrm{d}y\mathrm{d}t \end{align*} and so if \(f\in{}L^{p}((0,\infty))\) for \(1\le{}p<\infty\) then observe that, since \(\phi\) has support in \([0,1]\), then \begin{align*} \int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)|\tilde{f}(y)|\mathrm{d}y\mathrm{d}t&= \int_{0}^{\infty}\!{}\!{}\!{}\int_{t}^{t+\epsilon} e^{-st}\phi_{\epsilon}(t-y)|\tilde{f}(y)|\mathrm{d}y\mathrm{d}t\\ &\le\frac{\|\phi\|_{L^{\infty}(\mathbb{R})}}{\epsilon} \int_{0}^{\infty}\!{}\!{}\!{}\int_{t}^{t+\epsilon}\!{} e^{-st}|\tilde{f}(y)|\mathrm{d}y\mathrm{d}t\\ &\le\frac{\|\phi\|_{L^{\infty}((0,\infty))}}{\epsilon{}s}\cdot(\epsilon)^{1-\frac{1}{p}} \|\tilde{f}\|_{L^{p}((0,\infty))}\\ &=\frac{\|\phi\|_{L^{\infty}((0,\infty))}}{\epsilon{}s}\cdot(\epsilon)^{1-\frac{1}{p}}\|f\|_{L^{p}((0,\infty))}<\infty. \end{align*}

A similar proof works for \(p=\infty\). Thus, we may invoke Fubini's Theorem to obtain \begin{align*} \mathcal{L}\{\tilde{f}_{\epsilon}\}(s)&= \int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)\tilde{f}(y)\mathrm{d}y\mathrm{d}t =\int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)\tilde{f}(y)\mathrm{d}t\mathrm{d}y\\ &=\int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\int_{0}^{\infty}\!{} e^{-s(t-y)}\phi_{\epsilon}(t-y)\mathrm{d}t\mathrm{d}y =\int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\int_{0}^{\infty}\!{} e^{-su}\phi_{\epsilon}(u)\mathrm{d}u\mathrm{d}y\\ &=\int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\int_{0}^{\epsilon}\!{} e^{-su}\phi_{\epsilon}(u)\mathrm{d}u\mathrm{d}y =\left(\int_{0}^{\epsilon}\!{}e^{-su}\phi_{\epsilon}(u)\mathrm{d}u\right) \int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\mathrm{d}y\\ &=\left(\int_{0}^{\epsilon}\!{}e^{-su}\phi_{\epsilon}(u)\mathrm{d}u\right) \mathcal{L}\{\tilde{f}\}(s) =\left(\int_{0}^{\epsilon}\!{}e^{-su}\phi_{\epsilon}(u)\mathrm{d}u\right) \underbrace{\mathcal{L}\{f\}(s)}_{=0}. \end{align*} Since \(s>0\) was arbitrary then we conclude that for each \(\epsilon>0\) and \(s>0\) that \(\mathcal{L}\{\tilde{f}_{\epsilon}\}(s)=0\). I claim that as \(\epsilon\to0^{+}\) then \(\tilde{f}_{\epsilon}\) converges almost everywhere to \(\tilde{f}\). Observe that \begin{equation*} \tilde{f}_{\epsilon}(t)-\tilde{f}(t)= \frac{1}{\epsilon}\int_{0}^{\infty}\!{}\phi\left(\frac{t-y}{\epsilon}\right) \bigl(\tilde{f}(y)-\tilde{f}(t)\bigr)\mathrm{d}y. \end{equation*} Thus, if \(t\) is a Lebesgue point of \(\tilde{f}\), which almost every point is, then we obtain by the Lebesgue Differentiation Theorem that \begin{equation*} \|\phi\|_{L^{\infty}(\mathbb{R})}\cdot\frac{1}{\epsilon} \int_{t-\epsilon}^{t}|\tilde{f}(y)-\tilde{f}(t)|\mathrm{d}y\to0. \end{equation*} Next we demonstrate that \(\tilde{f}_{\epsilon}\in{}C_{0}((0,\infty))\) for each \(\epsilon\). Observe that for \(0<t_{1}<t_{2}<\infty\) we have, if \(1\le{}p<\infty\), that \begin{align*} \frac{1}{\epsilon}\int_{0}^{\infty}\!{} \left|\phi\left(\frac{t_{2}-y}{\epsilon}\right)- \phi\left(\frac{t_{1}-y}{\epsilon}\right)\right||\tilde{f}(y)|\mathrm{d}y\\ &=\frac{1}{\epsilon}\int_{[t_{1}-\epsilon,t_{1}]\cup[t_{2}-\epsilon,t_{2}]}\!{} \left|\phi\left(\frac{t_{2}-y}{\epsilon}\right)- \phi\left(\frac{t_{1}-y}{\epsilon}\right)\right||\tilde{f}(y)|\mathrm{d}y\\ &\le\frac{\|\nabla\phi\|_{L^{\infty}(\mathbb{R})}|t_{2}-t_{1}|}{\epsilon^{2}} \int_{[t_{1}-\epsilon,t_{1}]\cup[t_{2}-\epsilon,t_{2}]}\!{} &\le\frac{\|\nabla\phi\|_{L^{\infty}(\mathbb{R})}|t_{2}-t_{1}|}{\epsilon^{2}}\cdot \bigl(2\epsilon\bigr)^{1-\frac{1}{p}}\|\tilde{f}\|_{L^{p}((0,\infty))}\\ &=\frac{\|\nabla\phi\|_{L^{\infty}(\mathbb{R})}|t_{2}-t_{1}|}{\epsilon^{2}}\cdot \bigl(2\epsilon\bigr)^{1-\frac{1}{p}}\|f\|_{L^{p}((0,\infty))}. \end{align*}
 * \tilde{f}_{\epsilon}(t)-\tilde{f}(t)|\le{}
 * \tilde{f}_{\epsilon}(t_{2})-\tilde{f}_{\epsilon}(t_{1})|&\le{}
 * \tilde{f}(y)|\mathrm{d}y\\