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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 $p.$

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    $\begingroup$ What do you mean by $L(\widetilde{y})(p_{0})=0\Leftrightarrow y=0\text{ on }(0,\frac{1}{2})$? That this equivalence holds for all $y\in L^{2}(0,1)$? $\endgroup$ Oct 12, 2021 at 13:06
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    $\begingroup$ Also, concerning "let $\widetilde{y}$ be its extension on $(0,\infty ).$". How exactly is this extension defined? $\endgroup$ Oct 12, 2021 at 13:09
  • $\begingroup$ Thank you for your answers. @Iosif Pinelis Yes there us an equivalence for all $y\in L^2(0,1)$ and the extension is the multiplication by the characteristic function of $(0,1)$. $\endgroup$
    – Gustave
    Oct 12, 2021 at 16:07

1 Answer 1

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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.

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  • $\begingroup$ Thank you sir for this great answer, I'm very greatful. Maybe I didn't pose my question in a good way. I have edited the question with some explanations about its origin. $\endgroup$
    – Gustave
    Oct 13, 2021 at 10:38
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    $\begingroup$ @Gustave : I am glad you liked the answer. You can then mark it accordingly. As for your edit, it is substantial. So, to keep things in good order and to give your edit more exposure, it is probably better to post the edit as a separate question. $\endgroup$ Oct 13, 2021 at 13:12
  • $\begingroup$ Thank you again for the answer. $\endgroup$
    – Gustave
    Oct 13, 2021 at 13:17

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