User:Tkojar/Sandbox/Lerch's Theorem
This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
This article needs to be tidied. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code. |
It has been suggested that this page be renamed. To discuss this page in more detail, feel free to use the talk page. |
Theorem
Let $f \in \map {L^p} {\openint 0 \infty, e^{-at} }$ for $1 \le p \le \infty$.
Let $\laptrans s = 0$ 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:
- $\ds 0 = \map {\laptrans f} s = \int_0^\infty e^{-s t} \map f t \rd t = -\int_0^\infty \paren {e^{-t} }^{s - 1} \map f {-\map \ln {e^{-t} } } \paren {-1} e^{-t} \rd t$
and so making the substitution $u = e^{-t}$ so that $\d u = -e^{-t} \rd t$ gives
- $\ds 0 = \int_0^1 u^{s - 1} \map f {-\map \ln u} \rd u$
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:
- $\ds \forall n \in \N_{>0}: 0 = \int_0^1 u^n \map g u \rd u$
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:
- $\ds \map {\tilde f_\epsilon} t = \int_0^\infty \map {\phi_\epsilon} {t - y} \map {\tilde f} y \rd y$
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 $\ds \int_\R \map \phi t \rd t = 1$.
Observe that:
\(\ds \map {\laptrans {\tilde f_{\epsilon} } } s\) | \(=\) | \(\ds \int_0^\infty e^{-s t} \map {\tilde f_\epsilon} t \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty e^{-s t} \int_0^\infty \map {\phi_\epsilon} {t - y} \map {\tilde f} y \rd y \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \int_0^\infty e^{-s t} \map {\phi_\epsilon} {t - y} \map {\tilde f} y \rd y \rd t\) |
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:
\(\ds \int_0^\infty \int_0^\infty e^{-s t} \map {\phi_\epsilon} {t - y} \size {\map {\tilde f} y} \rd y \rd t\) | \(=\) | \(\ds \int_0^\infty \int_t^{t + \epsilon} e^{-s t} \map {\phi_\epsilon} {t - y} \size {\map {\tilde f} y} \rd y \rd t\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\norm \phi_{\map {L^\infty} {\openint 0 \to} } } {\epsilon s} \cdot \epsilon^{1 - \frac 1 p} \norm {\tilde f}_{\map {L^p} {\openint 0 \to} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm \phi_{\map {L^\infty} {\openint 0 \to} } } {\epsilon s} \cdot \epsilon^{1 - \frac 1 p} \norm f_{\map {L^p} {\openint 0 \to} }\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \infty\) |
A similar proof works for $p = \infty$.
Thus, we may invoke Fubini's Theorem to obtain:
\(\ds \map {\laptrans {\tilde f_\epsilon} } s\) | \(=\) | \(\ds \int_0^\infty \int_0^\infty e^{-s t} \map {\phi_\epsilon} {t - y} \map {\tilde f} y \rd y \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \int_0^\infty e^{-s t} \map {\phi_\epsilon} {t - y} \map {\tilde f} y \rd t \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \map {\tilde f} y e^{-s y} \int_0^\infty e^{-s \paren {t - y} } \map {\phi_\epsilon} {t - y} \rd t \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \map {\tilde f} y e^{-s y} \int_0^\infty e^{-s u} \map {\phi_\epsilon} u \rd u \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \map {\tilde f} y e^{-s y} \int_0^\epsilon e^{-s u} \map {\phi_\epsilon} u \rd u \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\int_0^\epsilon e^{-s u} \map {\phi_\epsilon} u \rd u} \int_0^\infty \map {\tilde f} y e^{-s y} \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\int_0^\epsilon e^{-s u} \map {\phi_\epsilon} u \rd u} \map {\laptrans {\tilde f} } s\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\int_0^\epsilon e^{-s u} \map {\phi_\epsilon} u \rd u} \underbrace {\map {\laptrans f} s}_{=0}\) |
Since $s>0$ was arbitrary then we conclude that for each $\epsilon > 0$ and $s > 0$ that $\map {\laptrans {\tilde f_\epsilon} } s = 0$.
I claim that as $\epsilon\to 0^+$ then $\tilde f_\epsilon$ converges almost everywhere to $\tilde f$.
Observe that:
- $\ds \map {\tilde f_\epsilon} t - \map {\tilde f} t = \frac 1 \epsilon \int_0^\infty \map \phi {\frac {t - y} \epsilon} \paren {\map {\tilde f} y - \map {\tilde f} t} \rd y$
Thus, if $t$ is a Lebesgue point of $\tilde f$, which almost every point is, then we obtain by the Lebesgue Differentiation Theorem that
- $\ds \size {\map {\tilde f_\epsilon} t - \map {\tilde f} t} \le \norm \phi_{\map {L^\infty} \R} \cdot \frac 1 \epsilon \int_{t - \epsilon}^t \size {\map {\tilde f} y - \map {\tilde f} t} \rd y \to 0$
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 $1 \le p < \infty$:
\(\ds \size {\tilde f_\epsilon (t_2) - \tilde f_\epsilon (t_1) }\) | \(\le\) | \(\ds \frac 1 \epsilon \int_0^\infty \norm {\map \phi {\frac {t_2 - y} \epsilon} - \map \phi {\frac {t_1 - y} \epsilon} } \map {\tilde f} y \Vert \rd y\) |
|
|||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \epsilon \int_{ \closedint {t_1 - \epsilon} {t_1} \mathop \cup \closedint {t_2 - \epsilon} {t_2} } \size {\map \phi {\frac {t_2 - y} \epsilon} - \map \phi {\frac {t_1 - y} \epsilon} } \size {\map {\tilde f} y} \rd y\) | ||||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\norm {\nabla \phi}_{\map {L^\infty} \R} \size {t_2 - t_1} } {\epsilon^2} \int_{\closedint {t_1 - \epsilon} {t_1} \mathop \cup \closedint {t_2 - \epsilon} {t_2} } \size {\map {\tilde f} y} \rd y\) | ||||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\norm {\nabla \phi}_{\map {L^\infty} \R} \size {t_2 - t_1} } {\epsilon^2} \cdot \paren {2 \epsilon}^{1 - \frac 1 p} \norm {\tilde f}_{L^p \openint 0 \infty}\) | ||||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm {\nabla \phi}_{\map {L^\infty} \R} \size {t_2 - t_1} } {\epsilon^2} \cdot \paren {2 \epsilon}^{1 - \frac 1 p} \norm f_{L^p \openint 0 \infty}\) |
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:
- $0 = \ds \lim_{s \mathop \to \infty} s \map {\laptrans {\tilde f_\epsilon} } s = \lim_{t \mathop \to 0^+} \map {\tilde f_\epsilon} t$
where we have used that $\map {\laptrans {\tilde f_{\epsilon} } } s = 0$ for all $s > 0$.
We conclude that $\tilde f_{\epsilon}$ extends to $t = 0$ by defining $\map {\tilde f_{\epsilon} } 0 = 0$.
Next observe that by Hölder's Inequality for Integrals we have
- $\ds \size {\map {\tilde f_\epsilon} x} \le \int_0^\infty \map {\phi_\epsilon} {x - y} \size {\map {\tilde f} y} \rd y \le \paren {\int_0^\infty \map {\phi_\epsilon} {x - y} \size {\map {\tilde f} y}^p \rd y}^{\frac 1 p}$
and so by dominated convergence applied to $\map {\phi_\epsilon} {x - y} \size {\map {\tilde f} y}^p$ we have that
- $\ds \lim_{x \mathop \to \infty} \map {\tilde f_\epsilon} x = 0$
Since $\epsilon > 0$ was arbitrary we conclude that $\tilde f_\epsilon \in C_0 \openint 0 \to$ 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 $1$ for each $\epsilon > 0$ we have that $\tilde f_\epsilon \equiv 0$.
Since $\tilde f_\epsilon$ converges almost everywhere to $\tilde f$ then we conclude that $\map {\tilde f} t = 0$ for almost every $t \in \openint 0 \infty$.
Since $\map {\tilde f} t = \map f t$ for $t > 0$ then $\map f t = 0$ for almost every $t \in \openint 0 \infty$.
We conclude that $f$ is $0$ almost everywhere on $\openint 0 \infty$.
Step $3$: Now suppose that $f \in \map {L^p} {\openint 0 \to, e^{-at} }$ for $a \ge 0$ and $1 \le p \le \infty$ and $\map {\laptrans f} s = 0$ for $s > a$.
Observe that in this case $e^{-a t} \map f t \in L^1 \openint 0 \to$ and for $s > 0$ we have:
- $\map {\laptrans {e^{-a t} \map f t} } s = \map {\laptrans f} {s + a} = 0$
and so $e^{-a t} \map f t = 0$ for almost every $t \in \openint 0 \infty$.
Thus, $\map f t = 0$ for almost every $t \in \openint 0 \infty$.
$\blacksquare$
Source of Name
This entry was named for Mathias Lerch.