Or more generally, are Lfunctions injective in some strips $a<\Re(s)<b$, where $0\leq a<b \leq 1$?

5$\begingroup$ I doubt we can even have injectivity on a vertical line, let alone a whole strip. $\endgroup$– WojowuJun 30 '18 at 10:56
The Riemann zeta function function is not injective in any strip $a<\Re(s)<b$ with $\frac{1}{2}<a<b$.
For $b\leq 1$ this follows, for example, from Theorem 11.10 in Titchmarsh: The theory of the Riemann zeta function (see also the remarks after the theorem in the book). The theorem (which is probably due to Bohr) implies that for any nonzero complex number $c$ and for any sufficiently large positive number $T$, there are $\gg T$ points in the rectangle $a<\Re(s)<b$ and $0<\Im(s)<T$ such that $\zeta(s)=c$. The implied constant here depends on $a,b,c$.
Alternatively, still for $b\leq 1$, the same conclusion follows from the simpler Theorem 11.9 (which states that the values $\log(\sigma+it)$ are dense in $\mathbb{C}$ for any $\frac{1}{2}<\sigma<1$) in conjunction with the open mapping theorem.
For $b>1$, we can employ Theorem 11.8 (A) in a similar fashion to reach the required conclusion.

1$\begingroup$ This works for $a >1/2$, but of course is not true if the strip is entirely to the left of the critical line (when the values of $\zeta$ are big). Some modification of this result on $a$points is probably needed in this case. $\endgroup$– LuciaJul 1 '18 at 6:29

$\begingroup$ @Lucia: Yes, I noticed that and added $0<a<b<1$, which is not a loss of generality from the OP's condition. I also reduced to $\frac{1}{2}<a<b<1$, so that the quoted theorems in Titchmarsh are readily applicable. For $a=1<b$, one can apply Theorem 11.6. Thanks for your comments! $\endgroup$ Jul 1 '18 at 6:35

2$\begingroup$ Hi GH: When $a=1/4$ and $b=1/3$ how does the functional equation immediately imply this? I don't see that it follows at once  for any fixed value of $a$ there are only finitely many $a$points in such a strip. $\endgroup$– LuciaJul 1 '18 at 6:37

$\begingroup$ @Lucia: Hmm, you are right. Perhaps I got up too early and thought that $\zeta(1s)=\zeta(s)$. Let me update my response! $\endgroup$ Jul 1 '18 at 6:38

$\begingroup$ @Lucia: I removed now the upper bound on $b$, but not the lower bound on $a$. I think I got "Bohred" with this problem (sorry for the bad joke), I leave it to others to finish it! $\endgroup$ Jul 1 '18 at 6:48
I think the answer is negative for the $\zeta$function as well as for many $L$functions and even products of them because injectivity in such strips would contradict various universality theorems. More precisely for Riemann's $\zeta$function:
By symmetry it suffices to consider the halfstrip $1/2<\sigma<1$. Voronin's universality tells us that for any compactum $K$ on this halfstrip with connected complement, any nonvanishing $f(s)$ holomorphic on the interior of $K$ can be approximated arbitrarily well uniformly by a shift $\zeta(s+i\tau)$, $s=\sigma+it$, for some fixed $\tau>0$, i.e. $\forall\varepsilon>0\ \exists\tau>0: \zeta(s+i\tau)f(s)_K<\varepsilon$ (in fact, the amount of such $\tau$s has positive lower density). Now take a sufficiently noninjective $f$.
Similar results hold for many $L$functions and products thereof (just google "universality of Lfunctions" to find some surveys).
Maybe there are easier arguments against noninjectivity of the $\zeta$function on strips inside the critical strip.

4$\begingroup$ This argument gives you many $\tau$'s with corresponding values $\zeta(s+i\tau)$ within $\varepsilon$ distance of each other. But how do you get equal values? There are injective functions $\varepsilon$close to noninjective ones! $\endgroup$ Jul 1 '18 at 3:32

3$\begingroup$ @GHfromMO We can salvage this as follows: take a function $f$ which takes values with both positive and negative imaginary part and approximate it using zeta function. By continuity there must be a real value taken and this value must be also taken at a conjugate point. We can use this to prove that we can't have injectivity on vertical lines contained in the critical strip (as I have remarked in a comment to the question). $\endgroup$– WojowuJul 1 '18 at 16:08

$\begingroup$ @Wojowu: OK, fair enough. However, Voronin's universality theorem works in the halfplane $\Re(s)>1/2$ only. But I do agree that this way M.G.'s solution can be fixed. On the other hand, Voronin's theorem came after Bohr's theorem, using the ideas of Bohr. $\endgroup$ Jul 1 '18 at 19:31

1$\begingroup$ @GHfromMO Indeed, that's still an issue to be considered. $\endgroup$– WojowuJul 1 '18 at 19:33

$\begingroup$ @Wojowu: I will now give a +1 to M.G.'s response, thanks to your comment. $\endgroup$ Jul 1 '18 at 19:35