Let $y\in L^{2}(0,1)$ and let $\widetilde{y}$ be its extension on $(0,\infty
).$ Assume that there exist $p\_{0},p\_{1}\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,$ $p\_{0}\neq p\_{1}$ such that
\begin{eqnarray\*}
L(\widetilde{y})(p\_{0}) &=&0\Leftrightarrow y=0\text{ on }(0,\frac{1}{2}), \\
L(\widetilde{y})(p\_{1}) &=&0\Leftrightarrow y=0\text{ on }(\frac{1}{2},1).
\end{eqnarray\*}


What can I say about the problem $L(y)(p)=0\Rightarrow y=0?.$ What are the
additional assumptions that guarantee $y=0?.$ Is there any references about
this problem?. Thank you in advance?


**Edited:**


The problem comes from the following problem: We want to prove that
the unique solution to the following difference equation is the null one:
$$
au(x)+b\mathbf{1}\_{(0,\frac{1}{2})}(x)u(x+\frac{1}{2})+c\mathbf{1}\_{(\frac{1%
}{2},1)}(x)u(x-\frac{1}{2})=0,\text{ }x\in (0,1).
$$
Extending $u$ by zero outside $(0,1)$ and taking the Laplace transform yields
$$
a\int\_{0}^{1}e^{-px}u(x)dx+be^{\frac{p}{2}}\int\_{\frac{1}{2}%
}^{1}e^{-px}u(x)dx+ce^{-\frac{p}{2}}\int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,%
\text{ }p\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,
$$
that is
$$
\left( a+be^{\frac{p}{2}}\right) \int\_{\frac{1}{2}}^{1}e^{-px}u(x)dx+\left(
a+ce^{-\frac{p}{2}}\right) \int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,\text{ }%
p\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
$$
If we let for instance $p=\gamma +4n\pi i,n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+ce^{-\frac{\gamma }{2}}=0$ we get
$$
\left( a+be^{\frac{\gamma }{2}}\right) e^{-\gamma }\int\_{\frac{1}{2}%
}^{1}e^{-4n\pi ix}u(x)dx=0,
$$
which yields that $u=0$ on $(\frac{1}{2},1)$ if $a+be^{\frac{\gamma }{2}%
}\neq 0$ which is equivalent to $a^2-bc \neq 0$. With the same manner, by choosing this time $p=\delta +4n\pi
i,n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+be^{\frac{\delta }{2}}=0$ we get
$$
\left( a+ce^{-\frac{\delta }{2}}\right) e^{-\delta }\int\_{0}^{\frac{1}{2}%
}e^{-4n\pi ix}u(x)dx=0,
$$
so if $a+ce^{-\frac{\delta }{2}}\neq 0$ we get $u=0$ on $(0,\frac{1}{2})$, which is equivalent to $a^2-bc \neq 0$.


I don't know if this kind of reasoning is correct since the Laplace
transform of $u$ is zero on subintervals with different choices of $p.$



Laplace transform injectivity for different values of $p$

> 
> **Claim:** Assuming that $L$ is the Laplace transform, there is no complex $p\_0$ such that for all $y\in L^2(0,1)$ we have the implication $L(\tilde y)(p\_0)=0\implies y=0$ on $(0,1/2)$.
> 
> 
> 


So, your conditions can never be fulfilled, and therefore they imply any statement, be it true or false.


*Proof of the Claim:* Take any complex $p\_0=a+ib$, where $a$ and $b$ are real. Let $x\_1$ and $x\_2$ be the functions in $L^2(0,1/2)$ defined by
$$x\_1(t):=e^{-at}\cos bt,\quad x\_2(t):=e^{-at}\sin bt$$
for $t\in[0,1/2]$, so that $e^{-p\_0t}=x\_1(t)-i x\_2(t)$. Then there is a **nonzero** function $z\in L^2(0,1/2)$ (orthogonal to both $x\_1$ and $x\_2$) such that
$$\int\_0^{1/2}x\_1(t)z(t)\,dt=0=\int\_0^{1/2}x\_2(t)z(t)\,dt.\tag{1}$$
Then, letting $y\in L^2(0,1)$ be defined by the conditions $y:=z$ on $[0,1/2]$ and $y:=0$ on $(1/2,1]$, we have
$$L(\tilde y)(p\_0)=\int\_0^1 e^{-p\_0t}y(t)\,dt \\ 
=\int\_0^{1/2}x\_1(t)z(t)\,dt-i\int\_0^{1/2}x\_2(t)z(t)\,dt=0.$$


So, $L(\tilde y)(p\_0)=0$ but $y$ is not $0$ on $(0,1/2)$.


Thus, the Claim is proved.


