$\int_{0}^{\infty}e^{-st}h(t)dt=0 \Rightarrow h(t)=0.$

Question

Suppose $h(t)$ is continuous function and $\int_{0}^{\infty}e^{-st}h(t)dt=0 ~\forall~ s>s_{0}$, then prove that $h(t)=0$.

I know "if a function is continuous, non-negative or non-positive, and its integration is zero, then function must be zero", which is intuitively clear.

But here asked question is beyond my knowledge. Here, exponential function is doing some miracle, but how? Would you like to help me?

Answer 1

Here is an argument:

Technical modification. Fix any $s_1 > s_0$, and let

$$ H_0 (x) = \int_{0}^{x} e^{-s_1 t}h(t) \, dt \quad \text{and} \quad H(x) = e^{s_1 x}H_0 (x). $$

Then $H_0(x)$ is differentiable and $H_0(x) = o(1)$ as $x \to \infty$. Then for any $s > s_1$, we have

$$ \int_{0}^{R} e^{-st} h(t) \, dt = \left[ e^{-(s-s_1)t} H_0(t) \right]_{t=0}^{t=R} + (s - s_1) \int_{0}^{R} e^{-(s-s_1)t} H_0(t) \, dt. $$

Taking $R \to \infty$, we find that

$$ \int_{0}^{\infty} e^{-st} H(t) \, dt = \int_{0}^{\infty} e^{-(s-s_1)t} H_0(t) \, dt = 0 $$

as well. Moreover, we have an exponential bound $H(x) = o(e^{s_1 x})$. From now on, we work with $H$ instead of $h$.

Main argument. Fix any $s > \max\{s_1, 0\}$. The exponential bound says that $t \mapsto e^{-st}H(t)$ is integrable on $[0, \infty)$. Now for any polynomial $p(x) = \sum a_k x^k$, we have

$$ \int_{0}^{\infty} p(e^{-st})e^{-st}H(t) \, dt = \sum a_k \int_{0}^{\infty} e^{-(k+1)st}H(t) \, dt = 0. $$

Now let $\varphi$ be any continuous function supported on a compact subset of $(0, 1)$. By the Stone-Weierstrass theorem, we can approximate $\varphi$ by a polynomial w.r.t. the supremum norm. So

$$ \left| \int_{0}^{\infty} (p(e^{-st}) - \varphi(e^{-st})) e^{-st}H(t) \, dt \right| \leq \| p - \varphi\|_{\infty} \int_{0}^{\infty} e^{-st}|H(t)| \, dt. $$

shows that, by taking $p \to \varphi$, we have

$$ \int_{0}^{\infty} \varphi(e^{-st})e^{-st}H(t) \, dt = 0 $$

as well. Now the integrand of the LHS is zero outside some compact interval, hence by an easy modification of the fundamental theorem of calculus of variation shows that $H \equiv 0$. Consequently, $h \equiv 0$ as well.

Answer 2

The assumption is equivalent to $$ \int_0^\infty e^{-sx}\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{1} $$ for all $s\gt0$

Since $$ \begin{align} \int_0^\infty\left(e^{-x}-e^{-2x}\right)^n\,\mathrm{d}x &=\int_0^\infty e^{-(n-1)x}\left(1-e^{-x}\right)^n\,\mathrm{d}e^{-x}\\ &=\int_0^1t^{n-1}(1-t)^n\,\mathrm{d}t\\ &=\frac{n!(n-1)!}{(2n)!}\tag{2} \end{align} $$ define $$ f_n(x)=n\binom{2n}{n}\left(e^{-x}-e^{-2x}\right)^n\tag{3} $$ Then $$ \int_0^\infty f_n(x)\,\mathrm{d}x=1\tag{4} $$ Furthermore, for $x\gt0$, $$ 0\lt4\left(e^{-x}-e^{-2x}\right)\le1\tag{5} $$ and if $x\ne\log(2)$, then the right inequality is strict.

According to $(10)$ from this answer, $$ \binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}\tag{6} $$ Combining $(3)$, $(5)$, and $(6)$, we get that for $x\gt0$ $$ 0\lt f_n(x)\le\sqrt{\frac{n}\pi}\,\left[4\left(e^{-x}-e^{-2x}\right)\right]^n\tag{7} $$ so that outside any neighborhood of $\log(2)$, $f_n(x)$ eventually tends monotonically to $0$.

$(4)$ and $(7)$ imply that $\frac{\log(2)}\alpha f_n\left(\frac{\log(2)}\alpha x\right)$ is an approximation of $\delta(x-\alpha)$.

According to the assumption, for any $n$ and $\alpha$, $$ \int_0^\infty\frac{\log(2)}\alpha f_n\left(\frac{\log(2)}\alpha x\right)\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{7} $$ Since $h(x)$ is continuous, $(7)$ implies that $h(x)=0$ for $x\ge0$.