Convergence in semi-norms topology Assuming the series of functions $\{\phi_{n}(x)\}$ and the function $\phi(x)$ are in Schwartz space $S(R)$,
now my question is this,
Does $\lim_{n\to\infty} \phi_{n}(x)\to \phi(x)$ (in Hilbert space $L^2[-\infty,\infty]$)
imply $\lim_{n\to\infty}\phi_{n}(x)\to \phi(x)$ (in semi-norms topology of Schwartz space)?
 A: Certainly not because this would imply ${\scr S}=L^2$ (as $\scr S$ is dense in $L^2$ and complete with its Frechet topology).
A: With regard to the follow-up question in the comment: it is possible to make a countable projective family of Hilbert spaces $H^k$ whose projective limit is the Schwartz space (further, with Hilbert-Schmidt transition maps), e.g., by completing Schwartz functions (or test functions) with respect to norms $|f|^2_k=\langle (-\Delta+x^2)^k f,f\rangle$. The compactness of the resolvent gives an orthonormal basis of eigenfunctions, which are provably in the Schwartz space, and which are orthogonal in every $H^k$ (these are Hermite polynomials times a suitable Gaussian). 
Then it is true that $f\in H^k$ (with $k\ge 0$) has an expansion in $H^k$, and an expansion in $L^2$. Of course, the expansion in $H^k$ converges in $L^2$, and by uniqueness the coefficients must be the same. (But beware that the norms of the eigenfunctions vary depending on $k$, although they remain orthogonal.)
Thus, for $f\in H^\infty={\mathcal S}$, the $L^2$ expansion does converge in ${\mathcal S}$.
