Show that $0$ is the only distribution solution of $u''-u=0$ in $\mathcal{S}'(\mathbb{R})$ 
Show that $0$ is the only distribution in $\mathcal{S}'(\mathbb{R})$ that satisfies $u''-u=0$.

So, I have that for each $\varphi\in\mathcal{S}(\mathbb{R})$ $$0=\langle u''-u,\varphi\rangle = \langle u,\varphi''-\varphi\rangle.$$
Any hints on how to proceed?
 A: Consider the map $\Phi: S(\mathbb{R})\to S(\mathbb{R})$, $\Phi(\varphi):=\varphi''-\varphi$. It is readily seen that $\Phi$ defines an isomorphism: Either, by explicitly working out the inverse using the theory of linear ordinary differential equations or, more elegantly, by means of the Fourier transform: In fact, $\Phi$ is a Fourier multiplier with symbol $m(\xi):=-(|\xi|^2+1)$, $\xi \in \mathbb{R}$, i.e., we have $\Phi=\mathcal{F^{-1} \circ m \circ \mathcal{F}}$, where $\mathcal{F}$ denotes the Fourier transform. Hence, $\Phi$ is an isomprphism being an composition of three isomorphisms (the Fourier transform and the mutliplication by a polynomial define isomorphisms on $S(\mathbb{R}))$. Thus, w have for all $\varphi \in S(\mathbb{R})$
$0=\langle u''-u,\varphi\rangle=\langle u,\varphi''-\varphi\rangle=\langle u,\varphi''-\varphi\rangle=\langle u,\Phi(\varphi)\rangle$ , which by the surjectivity of $\Phi$ implies
$0=\langle u, \psi\rangle$ for all $\psi\in S(\mathbb{R})$. Hence, $u=0$.
Remark: The classical solutions to the equation $u''-u=0$ is $u(x)=\alpha \cosh(x)+\beta \sinh(x)$ for some constants $\alpha, \beta\in \mathbb{R}$. This function is of exponential growth (unless $\alpha=\beta=0$), so it is not tempered and thus does not define an element of $S'(\mathbb{R})$.
A: To my mind, it is clearer if we grant ourselves that Fourier transforms "work" on tempered distributions (without necessarily invoking the literal definition via duality to Schwartz functions).
So, assuming $u$ is a tempered distribution (so has a Fourier transform... that behaves as we expect), $u''-u=0$ gives (up to normalizing constant in Fourier transform) $-x^2\widehat{u}-\widehat{u}=0$ (as tempered distributions). That is, $(x^2+1)\widehat{u}=0$. The left-hand side is the (legitimate...) multiplication of tempered distributions by a smooth function all whose derivatives are of moderate growth. A crucial point is that the multiplicative inverse $1/(x^2+1)$ is still a smooth function all whose derivatives are of moderate growth. So the obvious symbol-manipulation that gives $\widehat{u}=0/(x^2+1)=0$ (and $u=0$) is really a proof (that the only solution is $u=0$).
