Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2009 / A / 1

The quotient norm isChanging the representative does not change the infimum. The triangle inequality and absolute homogeneity follow from those of the original norm. Vanishing means $x$ belongs to the closure of $Y$, hence to $Y$ because the subspace is closed. Thus this is a norm.

Use the absolutely convergent series criterion for completeness. If $\sum \Vert {\xi }_n{\Vert }_{X/Y}<\infty$, choose representatives $z_n$ of ${\xi }_n$ satisfying $\Vert z_n\Vert \le \Vert {\xi }_n{\Vert }_{X/Y}+2^{-n}$. The series $\sum z_n$ converges in the Banach space $X$ to some $z$. Since the quotient map is contractive, $\sum {\xi }_n$ converges to $z+Y$. Consequently the quotient is a Banach space.

A set is totally bounded if for every $\epsilon >0$ it can be covered by finitely many open balls of radius $\epsilon$.

Here is the needed separated-sequence argument, including the relevant Riesz lemma. For a proper closed subspace $Y$, choose $z\notin Y$, put $d=\mathrm{dist}(z,Y)>0$, and choose $y\in Y$ with $\Vert z-y\Vert <2d$. Then $w=(z-y)/\Vert z-y\Vert$ is a unit vector at distance greater than $1/2$ from $Y$. Starting with one unit vector, apply this to the span of the vectors already chosen. These spans are closed and proper because they are finite-dimensional while $X$ is infinite-dimensional. The resulting unit vectors satisfy $\Vert w_m-w_n\Vert >1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence the closed unit ball is not totally bounded.

Take $S=T(B_X)$. A bounded linear operator is compact exactly when the closure of this set is compact. In a complete metric space this is equivalent to its being totally bounded.

If $T$ is compact, choose a finite $\epsilon$-net $z_1,\dots ,z_m$ for $T(B_X)$ and let $Y$ be their span. Then $\mathrm{dist}(Tx,Y)<\epsilon$ for $\Vert x\Vert \le 1$, giving $\Vert {\pi }_{Y}T\Vert \le \epsilon$.

Conversely, fix $\delta >0$ and take $Y$ with $\Vert {\pi }_{Y}T\Vert \le \delta /4$. For every $x\in B_X$ choose $y_x\in Y$ with $\Vert Tx-y_x\Vert <\delta /2$. These approximants satisfy $\Vert y_x\Vert \le \Vert T\Vert +\delta /2$, so form a bounded subset of a finite-dimensional space and have a finite $\delta /2$-net. The same centers give a finite $\delta$-net for $T(B_X)$. Thus it is totally bounded and $T$ is compact. In particular