User:Tkojar/Sandbox/Lerch's Theorem

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:

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:

A similar proof works for $p = \infty$.

Thus, we may invoke Fubini's Theorem to obtain:

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 $0 0$ 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 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 $t \in \openint 0 \infty$.