Composition of a distribution with a map

Suppose $\lambda \in C_c^\infty(\mathbb{R})^*$ is a distribution and $f: \mathbb{R} \to \mathbb{R}$ is a continuous map of the real line. In addition assume $f$ has compact support. How can I make sense of $f\circ \lambda$? I would like to define it by $$\langle \phi, f\ \circ \lambda \rangle := \langle \phi \circ f, \lambda\rangle \text{, for } \phi \in C^\infty_c(\mathbb{R}),$$ but $\phi \circ f$ is only continuous becasue $f$ is only assumed to be continuous.

My question is:

Does it makes sense to define $\langle \phi \circ f, \lambda\rangle$ through approximation of $f$ by smooth functions $f_i$? That is: $$\langle \phi \circ f, \lambda\rangle =\lim_{i \rightarrow \infty} \langle \phi \circ f_i, \lambda\rangle ?$$

-
Even for continuous $f$, this $f\circ \lambda$ does not define a distribution in general. If we set $f(x)=0$ for all $x$ then $\phi\circ f$ can be a non-zero constant function, but test functions should decay to infinity. – Colin McQuillan Dec 8 '12 at 11:46
@ColinMcQuillan, Suppose $f \in C_0(\mathbb{R})$ is a continuous function that goes to $0$ at infinity, or even better has compact support. – Cantor Dec 8 '12 at 12:16
my previous example ($f\equiv 0$) satisfies those conditions. – Colin McQuillan Dec 8 '12 at 19:48
@ColinMcQuillan, I see, you are right. At first, I thought this might be an artificial problem, but now I don't see how to get around it. So even if $f$ is a smooth function I will have the same problem? – Cantor Dec 8 '12 at 21:35

Your "compositional" notation seems odd to me, but anyway, from your formula I understood that you mean just the image of a distribution under that map. So, for example, the derivative of $\delta$ should be mapped like a tangent vector, which means that all sorts of bad things happen when you try to approximate a non-differentiable map (or a differentiable one, but without convergence of derivatives). The limit may not exist, or it can be anything, and it depends on the way of approximation.
The notation $f \circ \lambda$ stands for what is defined after; just a definition. I don't understand your comment about the derivative of $\delta$ being mapped like a tangent vector. What do you mean by this? – Cantor Dec 8 '12 at 12:22
I think I see what you mean. With my definition I would have a problem with derivatives of distributions because according to my definition: $\langle \phi \circ f, \lambda' \rangle = \lim \langle (\phi \circ f_i)', \lambda \rangle$, but the $f_i \rightarrow f$ is only in $C^0$ and so the derivatives $(\phi \circ f_i)'$ may not converge to $(\phi \circ f)'$. Is this what you mean? – Cantor Dec 8 '12 at 12:33
@Cantor: Well, yes, and the delta function is just an example that illustrates this. If, say, $f(0)=0$, than $f \circ \delta^\prime = f^\prime(0) \delta^\prime$, so your approximation won't converge unless the derivatives converge. The same applies to higher derivatives, so $C^\infty$ approximation seems "the" natural thing here. – Alexander Shamov Dec 8 '12 at 13:29