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 $p.$