Suppose we have 2 sequences $(f_n)$ and $(g_n)$ in Hilbert space $\mathcal{H}$ that are linearly independent and bounded. Is it known under what conditions there is a bounded linear operator $A \colon \mathcal{H} \to \mathcal{H}$ such that $$Af_n=g_n$$ for all n?
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This is not possible under your general assumptions. Consider the Hilbert space $H=l^2$. Set $g_n=e_n$, $e_n$ the unit sequences. Set $f_1=e_1$ and $f_n=e_1 + n^{-1} e_n$ for $n\ge2$. Assume that there is such a linear mapping with $Af_n=g_n$ for all $n$. Then $A$ is unbounded: Take $n\ge2$ and compute $$ Ae_n = A(nf_n-ne_1) = n (e_n-e_1). $$
If one assumes that $(f_n)$ is an orthonormal sequence and $\sum_{n=1}^\infty \|g_n\|^2<\infty$ then it works. Set $$ Ax:=\sum_{n=1}^\infty \langle x,f_n\rangle g_n, $$ which has the desired mapping properties. Moreover, $$ \|Ax\|\le \sum_{n=1}^\infty |\langle x,f_n\rangle|\cdot \| g_n\| \le \left(\sum_{n=1}^\infty |\langle x,f_n\rangle|^2\right)^{1/2}\left(\sum_{n=1}^\infty \| g_n\|^2 \right)^{1/2} \\ \le \|x\| \left(\sum_{n=1}^\infty \| g_n\|^2 \right)^{1/2}. $$
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Thank you for the reply. I am aware that such operator $A$ won't exist in majority of cases. The question was more about characterization of existence of $A$ in terms of $(f_n)$ and $(g_n)$. Still, your reply was very helpful :) – Behemont May 12 '17 at 18:45