Let $f$ be a temperate distribution. Suppose that $f$ is a solution to the equation
$ f'-f= \delta_0 +1 $.
I want to find $ \hat{f}$...
Here's what I did:
Since $ f'-f= \delta_0 +1 $, then $ \hat{f'}-\hat{f}= \hat{\delta_0} +\hat{1} $.
So, we have $ \hat{f'}(\xi)-\hat{f}(\xi)= \hat{\delta_0}(\xi) +\hat{1}(\xi) $,
i.e., $ \xi \hat{f } (\xi)-\hat{f}(\xi)= 1 +\delta_0(\xi) $,
i.e., $ \hat{f } (\xi) = \dfrac{1 +\delta_0(\xi) }{ \xi -1}$.
Is this correct? and what happens when $ \xi=1$?
Moreover, how can I show that the solution to the above equation in the space of temperate distribution is unique?
Thank you!
(I would usually post this as a comment, but apparently I need an account for that.)
Suppose that $f \in \mathscr{S}'(\mathbb{R}) $ is a solution to the equation $$ f'-f = \delta_0 + 1. $$
Since you are following Gerd Grubb's textbook, have a look at the technique described on page 108, starting around equation (5.41). We can set the differential operator $$P(D) = iD - 1 = \partial - 1,$$ noting that $D = \frac{1}{i}\partial$ (see page 424). Then the differential equation above can be written $$P(D)f = \delta_0 + 1.$$ (Quick sanity check: $\delta_0+1$ is indeed in $\mathscr{S}'(\mathbb{R})$, so we are OK to use this method.) Then applying Fourier transform (cf. (5.43)), we get $$(i\xi-1)\hat{u}(\xi) = \mathscr{F}[\delta_0+1](\xi) \overset{(5.37),(5.38)}{=} 1 + 2\pi\delta_0(\xi),$$ i.e., $$\hat{u}(\xi) = \frac{1+2\pi\delta_0(\xi)}{i\xi-1}.$$ This is well-defined in $\mathscr{S}'(\mathbb{R})$, since the denominator cannot be zero when $\xi \in \mathbb{R}$. I think this is essential.
For uniqueness, I believe it suffices to use the fact that the Fourier transform $\mathscr{F}$ is a homeomorphism from $\mathscr{S}'(\mathbb{R})$ onto itself (this is Theorem 5.17 in Grubb's book). Then the argument goes as follows: Suppose that $g \in \mathscr{S}'(\mathbb{R})$ is another solution to the above equation. Then $\hat{g}=\hat{f}$ by the above computation we have just done, so we find that $g = \mathscr{F}^{-1}[\hat{g}] = \mathscr{F}^{-1}[\hat{f}] = f$. Hence $f$ is a unique solution in $\mathscr{S}'(\mathbb{R})$.
Good luck tomorrow, right? :-)