Discontinuous function extended as a distribution

36 Views Asked by At

Let $f:\mathbb{R}^\star\to\mathbb{C}$ be a continuous function on its domain, and suppose : $$\exists m\in\mathbb{N}^\star,\;\exists c>0\;/\;\forall x\in[-1,1]\setminus\lbrace0\rbrace,\;|f(x)|\leqslant\frac{c}{|x|^m}$$

The goal is to prove that there exists a distribution $T_f\in\mathcal{D}'(\mathbb{R})$ that extends $f$ to $\mathbb{R}$, in the sense that if $\varphi\in\mathcal{D}(\mathbb{R}^\star)$ is a test function whose support does not contain $0$, then $<T_f,\varphi>=<f,\varphi>$.

Knowing that the principal values are given as : $$<\text{vp}\left(\frac{1}{x}\right),\varphi>=\lim_{\varepsilon\to0^+}\int_{|x|>\varepsilon}\frac{\varphi(x)}{x}\text{d}x$$ and : $$<\text{vp}\left(\frac{1}{x^2}\right),\varphi>=\lim_{\varepsilon\to0^+}\left(\int_{|x|>\varepsilon}\frac{\varphi(x)}{x^2}\text{d}x-\frac{2\varphi(0)}{\varepsilon}\right),$$ I would be tempted to look for $T_f$ as something like : $$<T_f,\varphi>=\lim_{\varepsilon\to0^+}\left(\int_{|x|>\varepsilon}f(x)\varphi(x)\text{d}x-\beta_\varepsilon(\varphi)\right)$$

Does anyone know something about this ? I can find no litterature, and I have no clue what the $\beta_\varepsilon(\varphi)$ would be besides a sum of coefficients of the form $\displaystyle\frac{\lambda\varphi^{(k)}(0)}{\varepsilon^\ell}$.

Thanks in advance for your suggestions and pro tips !