Codimension 1 closed subspace as a kernel of a functional

Question

My non-linear analysis book says that if I have a linear operator $T:X\to Y$ with close range $R$ and $\operatorname{codim}(R)=1$ (and also $\dim(\ker(T))=1$) then there exists $\phi\in Y^{*}$ such that $R=\{ y\in Y:\phi(y)=0\}$.

I suspect Hahn-Banach involved but I don't know why it is true. Could someone help me out?

Thank you.

No need for Hahn-Banach. If $V \subset Y$ is closed and has codimension $1$, what does that imply about $Y/V$? — Daniel Fischer, Jun 29 '15 at 17:55
I can decompose Y to V and Y/V as direct sum but where the functional comes from? — Lin, Jun 29 '15 at 18:13

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

Since the codimension of $R$ is one, the quotient space $Y/R$ has dimension one. (For quotients of Banach spaces, see this earlier MSE question, for example.) Since $Y/R$ has dimension one, there is a linear isomorphism $f:Y/R\to\mathbb R$ (assuming you are working over the reals). The quotient map $p:Y\to Y/R$ is linear and continuous and its kernel is $R$. Now you can set $\phi=f\circ p:Y\to\mathbb R$. Can you show that $\phi$ is continuous and find its kernel?

Codimension 1 closed subspace as a kernel of a functional

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