User:Tkojar/Sandbox/Lerch's Theorem

Theorem
Let $f \in L^p (\openint 0 \infty, e^{-at})$ for $1 \le p \le \infty$ and $\laptrans s = 0$ for all $s > a$.

Then $f = 0$ almost everywhere on $\openint 0 \infty$.

Proof
We prove this theorem in steps.

First we demonstrate this for $f \in C_0 (\openint 0 \infty)$.

Then we use the previous case to demonstrate the theorem for $f \in L^p (\openint 0 \infty)$ for $1 \le p \le \infty$.

Finally, we use the previous step to demonstrate this for $f \in L^p (\openint 0 \infty, e^{-a t})$.

Step $1$: Let $f \in C_0 (\openint 0 \infty)$.

Let $\map {\laptrans f} s$ for all $s > 0$.

Note that the Laplace transform is defined since $f \in L^p (\openint 0 \infty)$ for all $1 \le p \le \infty$.

Observe that:
 * $\displaystyle 0 = \map {\laptrans f} s = \int_0^\infty e^{-s t} \map f t \rd t = -\int_0^\infty \paren {e^{-t} }^{s - 1} \map f {-\map \ln {e^{-t} } } \paren {-1} e^{-t} \rd t$

and so making the substitution $u = e^{-t}$ so that $\d u = -e^{-t} \rd t$ gives


 * $\displaystyle 0 = \int_0^1 u^{s - 1} \map f {-\map \ln u} \rd u$

Observe that since $f \in C_0 (\openint 0 \infty)$ then $\map g u = \map f {-\map \ln u}$ extends to a continuous function defined on $\closedint 0 1$ by defining $\map g 0 = 0 = \map g 1$.

In particular, we have by choosing $s = 1, 2, 3, 4, \ldots$ that:
 * $\forall n \in \N_{>0} \displaystyle 0 = \int_0^1 u^n \map g u \rd u$

where the integral is understood as over the compact interval $\closedint 0 1$.

By the Weierstrass Approximation Theorem we obtain that $g \equiv 0$.

Thus, since $-\ln u$ is a bijection between $\openint 0 1$ and $\openint 0 \infty$, we have that $f \equiv 0$.

Step $2$: Now suppose $f \in L^p (\openint 0 \infty)$ for $1 \le p \le \infty$.

We extend $f$ to a function $\tilde f: \R \to \R$ by defining:
 * $\map {\tilde f} t = \begin{cases} \map f t & : t > 0 \\ 0 & : t \le 0 \end{cases}$

Now we define $\tilde f_\epsilon: \openint 0 \infty \to \R $, for $\epsilon>0$, by


 * $\displaystyle \map {\tilde f_\epsilon} t = \int_0^\infty \map {\phi_\epsilon} {t - y} \map {\tilde f} y \rd y$

where $\map {\phi_\epsilon} t = \dfrac 1 \epsilon \map \phi {\frac t \epsilon}$ and $\phi: \R \to \R$ is a non-negative, smooth, function supported in $\closedint 0 1$ such that $\displaystyle \int_\R \map \phi t \rd t = 1$.

Observe that: \begin{align*} \map {\laptrans {\tilde f_{\epsilon} } } s &= \int_{0}^{\infty}e^{-st} \map {\tilde f_\epsilon} t \rd t = \int_{0}^{\infty}e^{-st} \int_{0}^{\infty}\phi_{\epsilon}(t-y)\tilde{f}(y) \rd y \rd t\\ &=\int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)\tilde{f}(y) \rd y \rd t \end{align*}

and so if $f\in{}L^{p}(\openint 0 \infty)$ for $1\le{}p<\infty$ then observe that, since $\phi$ has support in $\closedint 0 1$, then

\begin{align*} \int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)|\tilde{f}(y)| \rd y \rd t&= \int_{0}^{\infty}\!{}\!{}\!{}\int_{t}^{t+\epsilon} e^{-st}\phi_{\epsilon}(t-y)|\tilde{f}(y)| \rd y \rd t\\ &\le\frac{\|\phi\|_{L^{\infty}(\R)}}{\epsilon} \int_{0}^{\infty}\!{}\!{}\!{}\int_{t}^{t+\epsilon}\!{} e^{-st}|\tilde{f}(y)| \rd y \rd t\\ &\le\frac{\|\phi\|_{L^{\infty}(\openint 0 \infty)}}{\epsilon{}s}\cdot(\epsilon)^{1-\frac{1}{p}} \|\tilde{f}\|_{L^{p}(\openint 0 \infty)}\\ &=\frac{\|\phi\|_{L^{\infty}(\openint 0 \infty)}}{\epsilon{}s}\cdot(\epsilon)^{1-\frac{1}{p}}\|f\|_{L^{p}(\openint 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) \rd y \rd t =\int_{0}^{\infty}\!{}\!{}\!{}\int_{0}^{\infty}\!{} e^{-st}\phi_{\epsilon}(t-y)\tilde{f}(y) \rd t \rd y\\ &=\int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\int_{0}^{\infty}\!{} e^{-s(t-y)}\phi_{\epsilon}(t-y) \rd t \rd y =\int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\int_{0}^{\infty}\!{} e^{-su}\phi_{\epsilon}(u) \rd u \rd y\\ &=\int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy}\int_{0}^{\epsilon}\!{} e^{-su}\phi_{\epsilon}(u) \rd u \rd y =\left(\int_{0}^{\epsilon}\!{}e^{-su}\phi_{\epsilon}(u) \rd u\right) \int_{0}^{\infty}\!{}\tilde{f}(y)e^{-sy} \rd y\\ &=\left(\int_{0}^{\epsilon}\!{}e^{-su}\phi_{\epsilon}(u) \rd u\right) \mathcal{L}\{\tilde{f}\}(s) =\left(\int_{0}^{\epsilon}\!{}e^{-su}\phi_{\epsilon}(u) \rd 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:


 * $\displaystyle \map {\tilde f_\epsilon} t - \map {\tilde f} t = \frac 1 \epsilon \int_0^\infty \map \phi {\frac {t - y} \epsilon} \paren {\map {\tilde f} y - \map {\tilde f} t} \rd y$

Thus, if $t$ is a Lebesgue point of $\tilde f$, which almost every point is, then we obtain by the Lebesgue Differentiation Theorem that


 * $\displaystyle \size {\map {\tilde f_\epsilon} t - \map {\tilde f} t} \le \norm \phi_{\map {L^\infty} \R} \cdot \frac 1 \epsilon \int_{t - \epsilon}^t \size {\map {\tilde f} y - \map {\tilde f} t} \rd y \to 0$

Next we demonstrate that $\tilde f_\epsilon \in C_0 (\openint 0 \infty)$ for each $\epsilon$.

Observe that for $0<t_{1}<t_2 < \infty$ we have, if $1 \le p < \infty$:

\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)| \rd 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)| \rd y\\ &\le\frac{\|\nabla\phi\|_{L^{\infty}(\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}(\R)}|t_{2}-t_{1}|}{\epsilon^{2}}\cdot \bigl(2\epsilon\bigr)^{1-\frac{1}{p}}\|\tilde{f}\|_{L^{p}(\openint 0 \infty)}\\ &=\frac{\|\nabla\phi\|_{L^{\infty}(\R)}|t_{2}-t_{1}|}{\epsilon^{2}}\cdot \bigl(2\epsilon\bigr)^{1-\frac{1}{p}}\|f\|_{L^{p}(\openint 0 \infty)}. \end{align*}
 * \tilde f_\epsilon (t_{2})-\tilde f_\epsilon(t_{1})|&\le{}
 * \tilde{f}(y)| \rd y\\

We conclude that $\tilde f_\epsilon$ is a Lipschitz function on $\openint 0 \infty$ for each $\epsilon > 0$ if $1 \le p < \infty$.

Note that a similar conclusion holds for $p = \infty$.

In particular, we note that $\tilde f_\epsilon$ is uniformly continuous on $\openint 0 \infty$ and hence $\tilde f_\epsilon$ extends to $0$.

Observe that by the initial value theorem we have:
 * $0 = \displaystyle \lim_{s \mathop \to \infty} s \map {\laptrans {\tilde f_\epsilon} } s = \lim_{t \mathop \to 0^+} \map {\tilde f_\epsilon} t$

where we have used that $\mathcal{L}\{\tilde{f}_{\epsilon}\}(s)=0$ for all $s>0$.

We conclude that $\tilde{f}_{\epsilon}$ extends to $t=0$ by defining $\tilde{f}_{\epsilon}(0)=0$.

Next observe that by H$\ddot{\text{o}}$lder's inequality we have

\begin{equation*} \end{equation*}
 * \bar{f}_{\epsilon}(x)|\le\int_{0}^{\infty}\!{}\phi_{\epsilon}(x-y)|\bar{f}(y)|\mathrm{d}y\le{}\left(\int_{0}^{\infty}\!{}\phi_{\epsilon}(x-y)|\bar{f}(y)|^{p}|\mathrm{d}y\right)^{\frac{1}{p}}

and so by dominated convergence applied to $\phi_{\epsilon}(x-y)|\bar{f}(y)|^{p}|$ we have that

\begin{equation*} \lim_{x\to\infty}\bar{f}_{\epsilon}(x)=0. \end{equation*}

Since $\epsilon>0$ was arbitrary we conclude that $\tilde{f}_{\epsilon}\in{}C_{0}((0,\infty))$ for all $\epsilon>0$.

By a similar proof, using that $\phi_{\epsilon}$ has compact support for each $\epsilon$, this conclusion holds also for $p=\infty$.

By step $1$ we have that for each $\epsilon>0$ that $\tilde{f}_{\epsilon}\equiv0$.

Since $\tilde{f}_{\epsilon}$ converges almost everywhere to $\tilde{f}$ then we conclude that $\tilde{f}(t)=0$ for almost every $t\in(0,\infty)$.

Since $\tilde{f}(t)=f(t)$ for $t>0$ then $f(t)=0$ for almost every $t\in(0,\infty)$.

We conclude that $f$ is $0$ almost everywhere on $(0,\infty)$. Step $3$: Now suppose that $f\in{}L^{p}((0,\infty),e^{-at})$ for $a\ge0$ and $1\le{}p\le\infty$ and $\mathcal{L}\{f\}(s)=0$ for $s>a$.

Observe that in this case $e^{-at}f(t)\in{}L^{1}((0,\infty))$ and for $s>0$ we have

\begin{equation*} \mathcal{L}\{e^{-at}f(t)\}(s)=\mathcal{L}\{f\}(s+a)=0 \end{equation*}

and so $e^{-at}f(t)=0$ for almost every $t\in(0,\infty)$.

Thus, $f(t)=0$ for almost every $t\in(0,\infty)$.