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This is a practice problem for an applied analysis qualifying exam.

If $\phi$ is any function defined on $\mathbb{R}^n$ and $\lambda>0$, let $\phi_\lambda(x) = \phi(\lambda x)$. We say that $\phi$ is homogenous of degree $\alpha$ if $\phi_\lambda = \lambda^\alpha \phi$ for any $\lambda>0$. If $T \in \mathcal{D}'(\mathbb{R}^n)$, we say that $T$ is homogenous of degree $\alpha$ if $T(\phi_\lambda) = \lambda^{-\alpha-n}T(\phi)$.

Define $T(\phi) = \int_{\mathbb{R}^n}f(x)\phi(x)dx \ \forall \ \phi \in \mathcal{D}(\mathbb{R}^n)$, where $f$ is a locally integrable function. Show that distribution $T$ is homogenous of degree $\alpha$ if and only if the function $f$ is homogenous of degree $\alpha$.

I have the $\Leftarrow$ direction of the problem (fairly straightforward), but am struggling with the forward direction. I tried using the fact that $\phi \in \mathcal{D}(\mathbb{R}^n)$ to reduce the integral to one over a compact $K = supp(\phi)$. As $f$ is locally integrable, it would then be integrable on $K$. But I got stuck after this--am I missing some key theorem? Or just some obvious step?

Chriz26
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1 Answers1

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According to your definition, the fact that $T$ is homogeneous means that for all $\lambda > 0$

$$\int f(x) \phi_\lambda (x) \ \Bbb d x = \lambda ^{-\alpha - n} \int f(x) \phi (x) \ \Bbb d x$$

or, equivalently,

$$\int f(x) \phi (\lambda x) \ \Bbb d x = \lambda ^{-\alpha - n} \int f(x) \phi (x) \ \Bbb d x .$$

Making the change of variable $y = \lambda x$ in the first integral, you get

$$\lambda^{-n} \int f(\lambda ^{-1} y) \phi (y) \ \Bbb d y = \lambda ^{-\alpha -n} \int f(y) \phi (y) \ \Bbb d y $$

or, equivalently,

$$\int \Big( f(\lambda^{-1}y) - \lambda^{-\alpha} f(y) \Big) \phi(y) \ \Bbb d y = 0 ,$$

and since this is true for every test function $\phi$, it follows that

$$f(\lambda^{-1}y) - \lambda^{-\alpha} f(y) = 0 .$$

Replacing $\lambda$ by $\lambda ^{-1}$ and rearranging gives

$$f(\lambda y) = \lambda ^\alpha f(y) .$$

Alex M.
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