Does CTR mode yield a PRF? Can the keystream generated by CTR mode be considered a pseudo-random function?

I give more information. Assume $f$ is a PRF. We define a function which takes 2 inputs, the desired length, and an arbirtrary value:

$F_l(r):=f(r\cdot l \cdot 1) \cdot f(r \cdot l \cdot 2) \cdot \cdots\cdot f(r\cdot l\cdot l-1)\cdot f(r \cdot l \cdot l)$ where $\cdot$ denotes the concatenate operation.

Is $F_l$ a PRF ?
It is not accurate to say that the keystream from AES-CTR is a pseudorandom function. However, it is a pseudorandom generator. Furthermore, the construction that you gave is close to working but it's unclear where the key fits in. I will therefore elaborate on what we can exactly say. 

Let $F$ be a pseudorandom function, and for simplicity assume that the key length, input length and output length are all the same (and of length denoted $n$). We denote by $F_k(x)$ the output of the PRF on input $x$, with key $k$. Then, we have the following two facts:


$G(k) = F_k(0)\cdot F_k(1) \cdot F_k(2)\cdots$ constitutes a pseudorandom generator.
$F'_k(x) = F_k(0 \cdot x) \cdot F_k(1 \cdot x) \cdot F_k(2 \cdot x) \cdots$ constitutes a pseudorandom function. Note that in this construction, $x$ has to be shorter than $n$ to provide "space" for the counter.


You want the first case when you want a PRG using AES. Especially when you have AES-NI this is a very fast option.

You want the second case when you want a PRF with input length $n$ and a larger output length. There are a number of ways of doing this. Another way is $F'_k(x) = G(F_k(x))$ where $G$ is any PRG.