 Solution

No. Consider the continuous even functionThe formulas agree at $|x|=e^{-1}$, and $f(x)\to 0$ at zero. Part (i) defines its even-weight principal-value distribution. However,If this were a regular distribution represented by $g\in L_{\mathrm{loc}}^1(\mathbb{R})$, testing on either side of zero would force $g=f/x$ almost everywhere there, by uniqueness of the locally integrable function representing a regular distribution. Changing the value at zero cannot remove the displayed divergence, contradicting local integrability of $g$.

More generally, the exact condition isNecessity is the preceding argument; sufficiency follows because $f/x$ is then a locally integrable function and its truncated integrals converge absolutely. The value $f(0)=0$ alone provides no such rate of decay.