= Solution
Assume real $k\ne0$, and put $a=|k|$. Away from the source point the <Green function> solves $G''-a^2G=0$. The two decay conditions select $G=Ae^{a(x-\xi)}$ on $x<\xi$ and $G=Be^{-a(x-\xi)}$ on $x>\xi$. Continuity at $\xi$ requires $A=B$: a jump in $G$ would create a derivative of the <Dirac delta> that is absent from the equation. Integrating the equation across the source gives $G'(\xi+)-G'(\xi-)=1$, so $-2aA=1$. Thus the negative of the <one-dimensional modified Helmholtz Green function> is
$$
\boxed{G(x;\xi)=-\frac{e^{-|k||x-\xi|}}{2|k|}.}
$$
Its <distributional derivative> verifies the delta equation, including its sign.
For a source for which the integral converges and the resulting function has the required decay, define
$$
\boxed{y(x)=\int_{\mathbb R}G(x;\xi)S(\xi)d\xi.}
$$
Applying the differential operator under the integral, or in the sense of <distributions>, gives $\mathcal Ly=\int\delta(x-\xi)S(\xi)d\xi=S(x)$. A continuous integrable source, for example, makes this convolution vanish at both infinities and gives a classical solution. If there were two solutions, their difference would be $Ae^{ax}+Be^{-ax}$; decay at positive infinity forces $A=0$, and at negative infinity forces $B=0$. Thus \b[the decaying solution is unique whenever it exists].
The assumptions needed for existence should not be dropped: an arbitrary source need not admit a decaying solution, as $S=1$ demonstrates. Also the displayed formula presupposes $k\ne0$; for $k=0$ no Green function for a point source can decay at both infinities, since its derivative must jump by one while both exterior linear pieces would have to be zero.
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