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Let $E,X,F$ be Banach spaces, let $S:E\to F$ and $T:X\to F$ be bounded linear operators, and assume $S$ is injective and $\mathrm{im}T\subseteq \mathrm{im}S$. Then $R=S^{-1}T:X\to E$ is a well-defined linear operator with closed graph. If $x_n\to x$ and $R{x}_n\to z$, continuity of both $S$ and $T$ givesInjectivity gives $z=Rx$. The closed graph theorem therefore makes $R$ bounded.^[ref]

It is not necessary for $S^{-1}$ to be bounded on its whole range. On ${\ell }^2$, take $S(x_n)=(x_n/n)$ and $T(x_n)=(x_n/n^2)$. Then $S^{-1}$ is unbounded, but $R(x_n)=(x_n/n)$ is bounded. The boundedness assumption on $T$ is essential to the closed-graph argument; it cannot be omitted from the hypotheses.

Let $X$ be a real Hausdorff topological vector space, and let $U$ be a bounded, convex, open neighbourhood of zero with $U=-U$. Its Minkowski functionalis a norm and $U=\{x:p_U(x)<1\}$. Convexity and symmetry give a real balanced set, so the gauge is a seminorm. Openness gives the displayed unit-ball identity.

To prove definiteness, suppose $p_U(x)=0$. For any zero neighbourhood $V$, boundedness of $U$ gives $\epsilon U\subseteq V$ for a sufficiently small $\epsilon >0$. Since $x\in \epsilon U$, it belongs to every zero neighbourhood. The Hausdorff property forces $x=0$. The same boundedness comparison shows that the norm topology agrees with the original topology.

For a complex vector space, require $U$ to be balanced under complex scalars; symmetry under minus one alone only supplies the real homogeneity. The argument is independent of finite dimension and requires no Euclidean inner or outer balls.^[ref]

Peter Philip authored notes on functional analysis beginning with topological vector spaces, seminorms and locally convex spaces. They develop foundational tools including Hahn-Banach separation. The progression explains how geometric separation and continuity remain available in function spaces where a single norm may not describe the natural topology.^[ref]