if we represents $\varphi$ as trgonometric polynomial, do we make the solution smoother?

Question

There is part in the article about inverse problem i could not understand. The integral operator S defined from $H^{\frac{1}{2}}(0,2\pi)\to L^2(0,2\pi)$ and $(S\varphi)(x)=f(x), x\in [0,2\pi]$. After applying Tikhonov method, it can be solved but in the article $\varphi$ is represented by trigonometric polynomial with degree $J=12$ and then Tikhonov method is applied. My question is why $\varphi$ is represented by trigonometric polynomial ? if we dont represent it with trigonometric polynomial, is the solution less smoother ? if we represents $\varphi$ as trgonometric polynomial, do we make the solution smoother ? Thanks for your help...