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Let us define a "scaling operator" $\hat{\Sigma}(\lambda)$ such that when it acts on any function $f(x)$, it gives $$\hat{\Sigma}(\lambda)f(x)=f(\lambda x).$$ Is it possible to represent the operator $\hat{\Sigma}(\lambda)$ as a differential (or integral) operator? If yes, what is it?

If $\phi(x)$ be an eigenfunction of $\hat{\Sigma}(\lambda)$, then $$\hat{\Sigma}(\lambda)\phi(x)=\phi(\lambda x)=s(\lambda)\phi(x).$$

From this, is it possible to find out how $s(\lambda)$ depends on $\lambda$ and the possible forms of $\phi(x)$?

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    What's the context here? "Any" function? Like on any domain, no regularity conditions, "any"? – 2017-01-01
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    @MartinArgerami For physics students like me, I mean, reasonably well-behaved functions (continuous, differentiable etc). – 2017-01-02

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The customary pseudodifferential realization of the dilation operator is $$ e^{(\ln \lambda) ~~x\partial_x} ~f(x) = f(\lambda x), $$ basically Lagrange's translation operator where you set $\ln x =y$.

The eigenfunctions of dilations are then pure powers, $$ \phi_\kappa(x) = c x^\kappa , $$ with $s(\lambda)=\lambda^\kappa$.