Proving a normed space is Banach

Question

Given Banach spaces $X$ and $Y$ and a surjective bounded linear map $T: X \to Y$, I'm having trouble figuring out how to show this norm gives a Banach space with $Y$: $$ |y|_n := \inf(|u|_X + n|v|_Y : u \in X, v \in V, Tu + v = y) \hspace{1em} \forall y \in Y. $$ I know that $|y|_n \leqslant n|y|_Y$ implies this new norm is equivalent to $|\cdot|_Y$, but for my purposes I want to prove it from definition of this norm. We have $$ |y_i - y_j|_n \to 0 \text{ as } \min(i,j) \to \infty $$ gives a sequence $$ |u_k|_X + n|v_k|_Y \to 0. $$ But, I'm stuck here. Can we get $$ |u_i - u_j|_X + n|v_i - v_j|_Y \to 0 $$ instead which would give $y = Tu_k + v_k \to Tu + v$.

Answer 1

A normed space $(Z,|\cdot|)$ is complete iff $\sum_{m=1}^\infty z_m$ is $|\cdot|$-convergent whenever $\sum_{m=1}^\infty |z_m|<\infty$. So assume $\sum_{m=1}^\infty |y_m|_n<\infty$. For each $m$, choose $u_m\in X$, $v_m\in Y$ such that $|u_m|_X+n|v_m|_Y<|y_m|_n+1/2^m$ and $y_m=Tu_m+v_m$. Then $\sum_{m=1}^\infty |u_m|_X<\infty$ and $\sum_{m=1}^\infty n|y_m|_Y<\infty$, and $u:=\sum_{m=1}^\infty u_m$, $v:=\sum_{m=1}^\infty y_m$ are convergent in $X,Y$, respectively, and $$\lim_N \Bigl|Tu+v-\sum_{m=1}^N y_m\Bigr|_n\leqslant \lim_N \Bigl|\sum_{m=N+1}^\infty u_m\Bigr|_X+n\Bigl|\sum_{m=N+1}^\infty v_m\Bigr|_Y=0.$$