I want to prove the following
Let $E,X$ be Banach spaces, and $Y\subset E$ a closed subspace with codimension $1$. Let $T:Y \to X$ be a continuous linear function. Then there exists a continuous linear extension $\bar{T}$ of $T$ defined on $E$
my attempt: Let's say that $E=Y \oplus \mathbb{R}e$, for some $e \in E\setminus{Y}$. Define for $r \in \mathbb{R} $ and $y \in Y$ $$ \bar{T}(y+re):=T(y) $$ so I tried to prove that $$ \sup_{x \in B_E} \|\bar{T}(x)\|=\sup_{\|y+re\|\leq 1} \|T(y)\|< \infty $$ I'd like to say something like this: There exists $M>0$ such that $$ \|y+re\|\leq 1\, \text{for some $r\in \mathbb{R}$}\Longrightarrow \|y\|\leq M $$ which, I suppose, comes from the fact that $Y$ is closed. If I had that, I could say the following $$ \sup_{\|y+re\|\leq 1} \|T(y)\|\leq \sup_{\|y\|\leq M}\|T(y)\| $$ and conclude the proof. Any hint would be welcome. Thanks in advanced.