Proof of Every Linear Operator on a Space with a Finite Orthonormal Basis is Bounded

Throughout, $[n]$ is the initial segment determined by $n$, and by the definition of a tuple an $n$-tuple in a set $X$ is a map from $[n]$ to $X$, so the finite sums below are formed from maps on $[n]$ as required.

Two preliminaries. First, if $a,b\in\mathbb{R}^{n}$ satisfy $a_{k}\le b_{k}$ for every $k\in[n]$, then $\sum_{k=1}^{n}a_{k}\le\sum_{k=1}^{n}b_{k}$. Indeed $0\le b_{k}-a_{k}$ for every $k$, so claim 5 of Properties of Finite Sums gives $0\le\sum_{k=1}^{n}(b_{k}-a_{k})$, while claims 2 and 3 of that lemma give

adding $\sum_{k=1}^{n}a_{k}$ to $0\le\sum_{k=1}^{n}b_{k}-\sum_{k=1}^{n}a_{k}$ gives the assertion. Second, if $x,y,c$ are real numbers with $x\le y$ and $0\le c$, then $0\le c(y-x)=cy-cx$ by the second order axiom of Ordered Field, so $cx\le cy$; we call this multiplying an inequality by a nonnegative number.

$0\le C$. Each $\lVert T(e_{k})\rVert$ is nonnegative by the positivity condition of the norm, so claim 5 of Properties of Finite Sums gives $0\le C$.

The bound. Let $u\in V$. Since $e$ is an orthonormal basis, claim 1 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions gives $u=\sum_{k=1}^{n}\langle e_{k},u\rangle e_{k}$. Applying $T$ and using claim 4 of Properties of Finite Sums of Vectors together with the homogeneity of the linear map $T$,

By Triangle Inequality for Finite Sums of Vectors and the absolute homogeneity of the norm, noting that the $n$-tuples in $\mathbb{R}$ with components $\lVert\langle e_{k},u\rangle T(e_{k})\rVert$ and $|\langle e_{k},u\rangle|\,\lVert T(e_{k})\rVert$ coincide and therefore have the same finite sum,

Each $e_{k}$ is a unit vector because $e$ is in particular orthonormal, so $\lVert e_{k}\rVert=1$, and claim 1 of The Induced Norm is a Norm, and Induces a Metric gives

Multiplying by the nonnegative number $\lVert T(e_{k})\rVert$ gives $|\langle e_{k},u\rangle|\,\lVert T(e_{k})\rVert\le\lVert u\rVert\,\lVert T(e_{k})\rVert$ for every $k\in[n]$, so by the first preliminary and then claim 3 of Properties of Finite Sums together with the commutativity of multiplication,

Transitivity of the order gives $\lVert T(u)\rVert\le C\,\lVert u\rVert$. Since $0\le C$, the operator $T$ satisfies the condition of Bound for a Linear Operator and Bounded Linear Operator and is therefore bounded.