Convention in Riesz representation theorem vs. tempered distribution theory We are working over the complex field here.  Sometimes analysis textbooks say that every continuous linear functional on $L^p$ is integration against some $f \in L^{p'}$ for $p\in (1, \infty)$, rather than replacing $f$ by its complex conjugate.  Of course, since this statement is existential, it is equivalent.  However, it is out of line with the definition of inner product in the case p=2, but that is only aesthetically unpleasing.  I worry that it becomes actually a mathematical problem when we move to other types of linear functionals.  For instance, when we view Schwartz space as a subset of the space of tempered distributions by thinking of Schwartz functions as integration functionals, we don't get consistency of basic operations on Schwartz functions with those operations on the corresponding tempered distribution if the conjugation is inserted.  But if you don't insert the conjugation, then the convention does not overlap with anything from viewing Schwartz as a subset of $L^2$.
I take it the latter has never forced a contradiction, and indeed I have never found one myself.  But is there a reason why this isn't so bad that can be easily spelled out, and can someone maybe even tell me why the pioneers of the field knew this inconsistency would be okay? (if difficult to keep track of)  When do you conjugate?
 A: Here is how I interpret your question; correct me if I am wrong.  Let $U$ be an open subset of Euclidean space equipped with Lebesgue measure.  A function $f \in L^q(U)$ is identified with a continuous linear functional on $L^p(U)$ (where $p$ and $q$ are conjugate exponents) via the formula
$$\langle f,g \rangle = \int_U g \overline{f}$$
Similarly, a locally integrable function $f$ is identified with a continuous linear functional on the space $D(U)$ of test functions via the formula
$$\langle f,g \rangle = \int_U g f$$
You are asking: why do we use the complex conjugate when describing the dual of $L^p(U)$ but not when describing the dual of $D(U)$?
Let me first comment that there is no inconsistency here: $L^p(U)$ and $D(U)$ are very different spaces, and it is rare that one would want to think of a function as defining a linear functional on both spaces simultaneously.  So the two conventions rarely (if ever) collide.  And even if they did, it would be pretty easy to change either convention at minimal cost to the theory.
So what is the reason for the different conventions?  The reason for the convention in the $L^p$ setting is to make the notation in general compatible with the important case $p=2$, wherein one wants $\langle, \rangle$ to be an inner product.  In order for $\sqrt{\langle f,f \rangle}$ to be a norm, the complex conjugate is required.  
There is, however, a cost associated with this convention: the pairing is conjugate-linear in the conjugated variable rather than linear.  This cost is pretty minimal because in fact all continuous linear functionals on $L^p$ come from functions in $L^q$ and thus it isn't really necessary to maintain separate notation for a function in $L^q$ and the linear functional on $L^p$ that it determines.  The analogous statement for the dual of $D(U)$ is far from true: there are many distributions which do not come from locally integrable functions, and thus it is useful to keep track of the difference between them.  But if we're thinking of distributions as functions on $D(U)$ then it is more notationally convenient to define them to be linear rather than conjugate linear.
