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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: $\langle \phi \circ f, \lambda\rangle =\lim_{i \rightarrow \infty} \langle \phi \circ f_i, \lambda\rangle ?$

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    @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?2012-12-08

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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.

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    For example, if the distribution acts continuously on $C^k$, which means that it is (not worse than) a $k$-th derivative of a measure, then you can compose it with $f \in C^k$, since the test functions are required to be just $C^k$-smooth.2012-12-08