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Added November 29, 2011: What follows is wrong; the answer seems rather to be yes.(See below for details.)

$\infty$ are discrete series.]

[Correction added Nov. 29, 2011: From the Galois/motivic point of view, we have an algebraic group (the Mumford--Tate group of some motive), with a representation (the particular motive), and the Mumford--Tate group contains a cocharacter whose eigenvalues are the Hodge numbers. Discrete series corresponds to the eigenspaces being one dimensional.We now apply some functoriality, which is essentially to say that we apply some multi-linear algebraic process to the representation. Now this can certainly produce eigenspaces for the cocharacter of multiplicity $> 1$. (E.g. the adjoint representation of $SL_3$ has a two-dimensional eigenspace.) So it seems that functoriality doesn't preserve being discrete series. It does preserve being tempered. And the remarks about not getting Artin motives still seem okay, since while the eigenspaces can become greater than $1$-dimensional, for all the eigenspaces to become trivial, we have to do something pretty destructive, like applying functoriality for the trivial representation.]

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[Added July 14, in response to Minhyong's question in the comments as to whether or not discrete series can convert into non-discrete series after applying some functoriality.The answer is essentially no, as I will now explain.

For an arithmetic geometer, one should think of an automorphic form on the adelic group$G(\mathbb A_K)$ as a morphism from the motivic Galois group (over the base number field $K$) to the $L$-group of $G$. (There are subtleties and caveats, of course, but they need not concern us here; all I will say about them is that automorphic forms can give rise to "motives" with non-integral $(p,q)$ in its Hodge decomposition, which necessitates enlarging the category of motives to an unknown larger category, whose hypothetical Tannakian group is called "the Langlands group".)

Now functoriality takes place when you have a map from the $L$-group of $G$ to the $L$-group of $H$; one can just compose this with a map from the motivic Galois group to the former, to obtain a map from the motivic Galois group to the latter. Functoriality is the assertionthat the corresponding automorphic form on $H(\mathbb A_K)$ exists.

Now given an automorphic form $\pi$, its factors at the primes $v|\infty$ encode (via the local Langlands corresondence for $\mathbb R$ or $\mathbb C$) the Hodge numbers of the corresponding motive. One feature of discrete series is that (among other properties) they give rise to regular Hodge numbers, i.e. to sequences of $h^{p,q}$ with each $h^{p,q}\leq 1$. Now our original automorphic rep'n $\pi$ on $G(\mathbb A)$ corresponds to a motive whose Mumford--Tate group lies in the $L$-group of $G$, and if $\pi$ has discrete series components at primes above $\infty$, it has regular Hodge numbers at every place dividing $\infty$. If we then pass to a new motive by applying some map from the $L$-group of $G$ to the $L$-group of $H$, then concretely this corresponds to doing some kind of multilinear algebra on our motive, and the only way this can kill the property of having regular Hodge--Tate weights is if we do something like taking the diagonal map from the $L$-group of $G$ into its product with itself, and then embed the latter into the $L$-group of $H$.All such constructions will necessarily be a "reducible" rep'n of the $L$-group of $G$in the $L$-group of $H$ (more precisely, the centralizer will be a non-trivial Levi),and the corresponding automorphic form won't be a cuspform.

But even if we destroy the property of having regular Hodge numbers, we typically stilldon't have an Artin motive. To get an Artin motive we have to have $h^{p,q} = 0$ unless$p = q = 0$, and to do this, we have to do even more destructive things, like map the $L$-group of $G$ into the $L$-group of $H$ via the trivial representation,or something similar.Again, this won't correspond to any kind of interesting automorphic forms, justthose that correspond to (certain) sums of characters. So we can't produce interestingGalois type automorphic forms out of automorphic forms whose factors at primes above$\infty$ are discrete series.]

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of $GL_n(Gal(\bar{K}/K)$. [Aside: Minhyong asked for a sketch of a proof of this; here goes: fixing the representation at infinity means that we are fixing a bunch of elliptic operatorsthat the automorphic forms must satisfy. Fixing the level means that we are working on some particular quotient $X/\Gamma$ (here $X$ is the symmetric space attached to the real groupin question, in our particular case $GL_n(K\otimes \mathbb R)$, and $\Gamma$ is the fixed level). This need not be compact (indeed won't be in our particular case), but the cuspidal condition (indeed, even the moderate growth condition that non-cuspidal automorphic formsare required to satisfy) means that we can pretend it is, since we explicitly rule out the possibility of extreme growth at infinity. So now we looking at sections of some bundle on a compact space satsifying a bunch of elliptic equations, and such a space of sections if finite dimensional. (The holomorphic modular forms case is the most familiar: in this case the elliptic equations are the Cauchy--Riemann equations. In the Maass form case, the corresponding fact is the finiteness of the eigenspaces of the Laplacian. These aregood models for the general case.)]

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