The graph $G(T)=\{(x,Tx):x\in X\}$ is closed if it is a closed subset of $X\times Y$. Equivalently, $x_n\to x$ and $T{x}_n\to y$ imply $Tx=y$.

The closed graph theorem states that an everywhere-defined linear map between Banach spaces is bounded if and only if its graph is closed. Completeness of both spaces is part of the theorem. A bounded map has a closed graph by continuity.

Conversely, if the graph is closed, it is a Banach space in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. The coordinate projection $P:G(T)\to X$, $P(x,Tx)=x$, is a bounded linear bijection. The open mapping theorem makes $P^{-1}$ bounded. The other coordinate projection $Q:G(T)\to Y$ is bounded, and $T=Q{P}^{-1}$ is therefore bounded. This proves the theorem from the permitted result.

Linearity and $\Vert Tf{\Vert }_{\infty }\le 2\pi \Vert f{\Vert }_{\infty }$ show that the integral defines a bounded linear operator. Inductively,For $k=0$ this is immediate. If it holds for $k$, then $|\sin s|\le 1$ gives $|T^{k+1}f(t)|\le {\int }_0^t{s}^k\Vert f{\Vert }_{\infty }/k!ds$, which proves the next case. Thus $\Vert T^k\Vert \le (2\pi {)}^k/k!$.

For every $\lambda \ne 0$, the series $A_{\lambda }={\sum }_{k\ge 0}{\lambda }^{-k-1}{T}^k$ converges in operator norm, dominated by $|\lambda {|}^{-1}{e}^{2\pi /|\lambda |}$. Multiplying its partial sums by $\lambda I-T$ on either side gives the identity minus a remainder ${\lambda }^{-N-1}{T}^{N+1}$, which tends to zero by the factorial estimate. Therefore $A_{\lambda }$ is the bounded inverse of $\lambda I-T$.

Finally every function in the image of $T$ vanishes at zero, so $T$ is not onto the continuous functions. Hence $0$ belongs to its spectrum and

Let $Z=\overline{\mathrm{span}}\{x_n:n\ge 1\}$. If $Z\ne X$, part (a) provides a nonzero bounded functional annihilating $Z$; normalize it to have norm one. Density of the sequence in the dual unit ball gives an index $n$ with $\Vert f_n-f\Vert <1/4$. Since $f(x_n)=0$ and $\Vert x_n\Vert \le 1$,But the hypothesis and reverse triangle inequality givea contradiction. Therefore the span is dense.

If $X^{\prime }$ is separable, its unit ball has a countable dense set. For each nonzero $f_n$, the definition of its norm permits choosing $x_n$ in the unit ball with $|f_n(x_n)|\ge \Vert f_n\Vert /2$; choose zero for a zero functional. The preceding result gives a dense countable span over $\mathbb{Q}$, or over $\mathbb{Q}+i\mathbb{Q}$ in the complex case. Thus $X$ is separable.

The converse fails: ${\ell }^1$ is separable, since finite rational-coordinate sequences are dense, but $({\ell }^1{)}^{\prime }\cong {\ell }^{\infty }$ is not separable. The dual identification sends a bounded sequence $a$ to $x\mapsto {\sum }_n{a}_n{x}_n$, isometrically; conversely a bounded functional is determined by its values on the coordinate vectors and hence has this form by truncation. The uncountable family of zero-one sequences in ${\ell }^{\infty }$ has pairwise distance one, which no countable dense set can approximate within a radius smaller than $1/2$.

For the centered ball, the Hahn-Banach theorem supplies a norm-one real functional with $f(y)=\Vert y\Vert >r$. Hence $f(z)\le r$ on the ball, and any $t\in (r,\Vert y\Vert )$ separates it strictly from $y$.

For the general center, apply the same construction to $y-x$: choose $\Vert f\Vert =1$ and $f(y-x)=\Vert y-x\Vert$. ThenFor example,gives $f(z)<t<f(y)$ for every $z\in B$. The strict gap uses exactly the assumption that $y$ lies outside the closed ball.

Put $d=\mathrm{dist}(x,Z)$. If $d=0$, the zero functional works. If $d>0$, then $x\notin \overline{Z}$, and on $M=Z+\mathrm{span}\{x\}$ defineThe representation is unique, so this is linear. For $a\ne 0$,and the bound also holds for $a=0$. Thus $\Vert f_0\Vert \le 1$. The Hahn-Banach theorem extends it to $f\in X^{\prime }$ with the same bound, provingNo closedness assumption on $Z$ was required; distance is unchanged on taking its closure.

The series ${\sum }_{n\ge 1}{x}_n$ converges in $X$ when its partial sums $s_N={\sum }_{n=1}^N{x}_n$ converge in norm to an element of $X$. It is absolutely convergent when ${\sum }_n\Vert x_n\Vert <\infty$.

If $X$ is complete, thenfor an absolutely convergent series. Its partial sums are Cauchy and therefore converge.

Conversely suppose every absolutely convergent series converges. Given a Cauchy sequence $(z_n)$, choose a subsequence $(z_{n_k})$ with $\Vert z_{n_{k+1}}-z_{n_k}\Vert \le 2^{-k}$. The seriesis absolutely convergent. Its partial sums are precisely the subsequence, which therefore converges to some $z\in X$. A Cauchy sequence with a convergent subsequence converges to that same limit: estimate $\Vert z_n-z\Vert$ by $\Vert z_n-z_{n_k}\Vert +\Vert z_{n_k}-z\Vert$. Hence