Functional analysis starts by putting a topology on linear algebra. The norm is not only a size function; it determines which linear maps, limits, quotients, and dual objects are analytically meaningful.
Normed linear spaces
This is where topology first enters linear algebra. The norm determines convergence, continuity, and the meaning of approximation.
Definition. Let \(\mathbb F\in\{\mathbb R,\mathbb C\}\). A norm on an \(\mathbb F\)-vector space \(X\) is a map \(\|\cdot\|:X\to[0,\infty)\) such that \(\|x\|=0\) iff \(x=0\), \(\|\alpha x\|=|\alpha|\|x\|\), and \(\|x+y\|\le \|x\|+\|y\|\). The pair \((X,\|\cdot\|)\) is a normed space. The metric is \(d(x,y)=\|x-y\|\), and the open balls are \(B(x,r)=\{y:\|y-x\|<r\}\).
Proposition. Addition \(X\times X\to X\) and scalar multiplication \(\mathbb F\times X\to X\) are continuous. Hence every normed space is a topological vector space.
Proof. For addition, \(\|(x_n+y_n)-(x+y)\|\le \|x_n-x\|+\|y_n-y\|\). For scalar multiplication, \(\|\alpha_n x_n-\alpha x\|\le |\alpha_n|\|x_n-x\|+|\alpha_n-\alpha|\|x\|\), and a convergent scalar sequence is bounded. These estimates are exactly the sequential form of continuity in metric spaces.
Operator norm
A bounded linear operator is one whose effect on the unit ball is uniformly controlled. The operator norm records the sharp such control.
Definition. For normed spaces \(X,Y\), a linear map \(T:X\to Y\) is bounded if there exists \(C<\infty\) such that \(\|Tx\|_Y\le C\|x\|_X\) for all \(x\in X\). The operator norm is \(\|T\|=\inf\{C:\|Tx\|\le C\|x\|\ \forall x\in X\}\). The space of bounded linear maps is denoted \(\mathcal L(X,Y)\).
Proposition. If \(T\in\mathcal L(X,Y)\), then \(\|T\|=\sup_{\|x\|\le1}\|Tx\|=\sup_{\|x\|=1}\|Tx\|\) when \(X\ne\{0\}\), and \(\|Tx\|\le \|T\|\,\|x\|\) for all \(x\).
Proof. Every admissible constant \(C\) satisfies \(\|Tx\|\le C\) on the closed unit ball, so the first supremum is at most the infimum of such \(C\). Conversely, if \(M=\sup_{\|x\|\le1}\|Tx\|\), then for \(x\ne0\), \(x/\|x\|\) belongs to the unit sphere and \(\|Tx\|=\|x\|\,\|T(x/\|x\|)\|\le M\|x\|\). Thus \(M\) is admissible. The unit ball and unit sphere suprema agree by homogeneity.
Continuity of linear maps
Linearity makes continuity a local question at the origin. In normed or locally convex spaces, this local control becomes a usable global estimate.
Theorem. For a linear map \(T:X\to Y\) between normed spaces, the following are equivalent: \(T\) is bounded; \(T\) is continuous at \(0\); \(T\) is continuous at one point; \(T\) is continuous everywhere; \(T\) is Lipschitz.
Proof. If \(T\) is bounded, then \(\|Tx-Ty\|\le \|T\|\,\|x-y\|\), so \(T\) is Lipschitz and continuous everywhere. Continuity everywhere implies continuity at any point, and continuity at one point \(x_0\) implies continuity at \(0\), since \(T(h)=T(x_0+h)-T(x_0)\).
Assume \(T\) is continuous at \(0\). Choose \(\delta>0\) such that \(\|u\|<\delta\) implies \(\|Tu\|<1\). If \(x\ne0\), put \(u=\delta x/(2\|x\|)\). Then \(\|u\|=\delta/2\), hence \(\|Tu\|<1\), and by homogeneity \(\|Tx\|<(2/\delta)\|x\|\). Thus \(T\) is bounded.
Equivalent norms
Equivalent norms give the same topology even if they measure vectors differently. In finite dimension this always happens; in infinite dimension it becomes a serious distinction.
Definition. Two norms \(\|\cdot\|_1,\|\cdot\|_2\) on \(X\) are equivalent if there exist constants \(0<c\le C<\infty\) such that \(c\|x\|_1\le \|x\|_2\le C\|x\|_1\) for all \(x\in X\).
Proposition. Equivalent norms induce the same convergent sequences, the same open sets, and the same continuous linear maps.
Proof. The inequalities give \(B_1(0,r/C)\subset B_2(0,r)\subset B_1(0,r/c)\). Translating balls gives equality of the two metric topologies. Since continuity is a topological property, the continuous linear maps are the same.
Theorem. On a finite-dimensional vector space, all norms are equivalent.
Proof. Fix a basis \(e_1,\ldots,e_n\) and write \(x=\sum_i a_i e_i\). For any norm \(\|\cdot\|\), \(\|x\|\le(\max_i\|e_i\|)\sum_i |a_i|\), so the coordinate map from \((\mathbb F^n,\|\cdot\|_1)\) to \((X,\|\cdot\|)\) is continuous. The function \(a\mapsto\|\sum_i a_i e_i\|\) is continuous on the compact \(\ell^1\)-unit sphere and has a positive minimum, because it vanishes only at \(a=0\). This yields \(m\sum_i|a_i|\le \|x\|\le M\sum_i|a_i|\). Comparing two arbitrary norms through the coordinate \(\ell^1\)-norm proves equivalence.
Quotient norm
The quotient norm measures distance to a subspace. It is the correct norm on cosets only when the subspace is closed.
Definition. If \(M\) is a closed linear subspace of a normed space \(X\), the quotient \(X/M\) is normed by \(\|x+M\|=\inf_{m\in M}\|x-m\|\).
Proposition. The quotient formula defines a norm on \(X/M\) exactly when \(M\) is closed. The quotient map \(q:X\to X/M\), \(q(x)=x+M\), is linear and bounded with \(\|q\|\le1\).
Proof. The formula is the distance from \(x\) to \(M\). It is homogeneous and subadditive because \(M\) is linear. If the distance is \(0\), then \(x\in\overline M\), hence \(x\in M\) when \(M\) is closed, so \(x+M=0\). Conversely, if \(M\) is not closed and \(x\in\overline M\setminus M\), then \(x+M\ne0\) but its quotient seminorm is \(0\). Finally, \(\|q(x)\|=\inf_{m\in M}\|x-m\|\le \|x\|\).
Discontinuous linear functionals
This example separates algebra from analysis. Infinite-dimensional spaces have algebraic linear functionals that ignore the norm topology completely.
Example. In infinite-dimensional normed spaces, linearity alone does not imply continuity. If \(X\) has a countable linearly independent sequence \(e_n\) with \(\|e_n\|=1\), extend it to a Hamel basis and define \(f(e_n)=n\) on this sequence and \(f=0\) on the other basis vectors. Then \(f\) is linear on \(X\), but \(|f(e_n)|=n\) on the unit sphere, so \(f\) is unbounded and therefore discontinuous. This example uses the algebraic basis theorem and is not constructive in separable analytic coordinates.