Past exam of the functional analysis I course of the University of Oxford / 2023 / 1 / b

Since $W$ is proper, $T$ is a nonzero linear functional. Choose $e\in X$ with $Te=1$, by rescaling any vector outside its kernel. For every $x\in X$,If $ae\in W$, applying $T$ gives $a=0$. Consequently this is a direct sum, and the quotient vector space $X/W$ has dimension one. This computes the codimension of a vector subspace:

To prove boundedness without assuming completeness, use the linear functional with closed kernel criterion directly. Since $W$ is a closed set and $e\notin W$, the distanceis positive. If $Tx\ne 0$, then $e-x/(Tx)\in W$, soFor $Tx=0$ the resulting estimate is immediate. ThereforeHere $X^{\ast }$ denotes the continuous dual space.

The bounded linear functional from part (i) satisfies $|Tx|\le \Vert T\Vert \Vert x{\Vert }_X$, where $\Vert T\Vert$ is its operator norm. HenceThese are equivalent norms. Applied to $x_n-x_m$, the inequalities show that a sequence is a Cauchy sequence for either norm exactly when it is one for the other. Applied to $x_n-x$, they show that the two norms give the same limits in $X$. Thus every Cauchy sequence has a limit in one norm precisely when it does in the other: