convergence divergence - Vector space of complex sequences of the form $a_{2n}=n²a_{2n-1}$ for $n=1,2,...$ is closed (making use of the $l_{infty}$ norm)

I am trying to show that the subspace, say $A$, of $c_0$ (which I use to denote the space of all complex sequences converging to $0$, equipped with the $l_{\infty}$ norm) which contains the sequences of the form $(a_n)$ with $a_{2n}=n^2a_{2n-1}$ for $n=1,2,...$ is closed.

I think I picture the reason why by taking a sequence of elements of $A$ and working by contradiction, but I can't get a proper proof.

After quite some work done with $\epsilon$'s all around, I get nothing neat... Could you help me by explaining how you would write this down?

Answer

Define a mapping $f:c_0 \to c_0$ by $f(\{a_i\}_{i=1}^\infty) = \{k^{-2} a_{2k} - a_{2k-1}\}_{k=1}^\infty$.

This map is bounded and linear, so continuous.

The kernel of $f$ is $A = \{ \{a_k\}_{k=1}^\infty \mid a_{2k} = k^2 a_{2k-1} \}$.

Therefore $A$, as the pre-image of a closed set $\{0\}$, is closed.