Find a function $g$ such that $(1*g)(t)=t^{-1/2}$.

Question

Is there some function $g: (0,+\infty) \to \mathbb{R}$ satisfying $$(1*g)(t)=t^{-1/2}$$ for all $t>0$? Here the sign $*$ represents the convolution. I tried apply the Laplace transform both site of the equality, but i arrive that $$g(t)=[\mathcal{L}^{-1}(\lambda^{\frac{1}{2}})](t)$$ so I can't develop it. Can somebody help me?