[proofplan] We verify the three norm axioms directly from the definition of the [operator norm](/page/Operator%20Norm). Nonnegativity is inherited from the norm on $Y$, and definiteness follows by applying the zero norm condition to a normalized nonzero vector of $X$. Absolute homogeneity follows from homogeneity in $Y$, while the triangle inequality follows from the triangle inequality in $Y$ and then taking a supremum over the closed unit ball of $X$. [/proofplan]

[step:Check that the operator norm is nonnegative and finite] Let $T\in\mathcal L(X,Y)$. By definition, \begin{align*} \|T\|_{\mathcal L(X,Y)}=\sup\{\|Tx\|_Y:x\in X,\ \|x\|_X\le 1\}. \end{align*} For every $x\in X$ with $\|x\|_X\le 1$, the norm axiom in $Y$ gives $\|Tx\|_Y\ge 0$. Hence the supremum is nonnegative. Since $T$ is a bounded [linear map](/page/Linear%20Map), there exists a constant $C\ge 0$ such that $\|Tx\|_Y\le C\|x\|_X$ for every $x\in X$. Therefore $\|Tx\|_Y\le C$ whenever $\|x\|_X\le 1$, so the defining supremum is finite. [/step]

[step:Prove definiteness by normalizing nonzero vectors]First let $0_{X,Y}:X\to Y$ denote the zero linear map. For every $x\in X$ with $\|x\|_X\le 1$, one has $0_{X,Y}x=0_Y$, and therefore $\|0_{X,Y}x\|_Y=0$. Hence \begin{align*} \|0_{X,Y}\|_{\mathcal L(X,Y)}=0. \end{align*} Conversely, assume $T\in\mathcal L(X,Y)$ satisfies $\|T\|_{\mathcal L(X,Y)}=0$. Let $x\in X$. If $x=0_X$, then $Tx=0_Y$ by linearity. If $x\ne 0_X$, define \begin{align*} u:=\frac{x}{\|x\|_X}\in X. \end{align*} Then $\|u\|_X=1$, so the definition of the supremum gives \begin{align*} 0\le \|Tu\|_Y\le \|T\|_{\mathcal L(X,Y)}=0. \end{align*} Thus $\|Tu\|_Y=0$, and definiteness of the norm on $Y$ gives $Tu=0_Y$. By linearity of $T$, \begin{align*} Tx=T(\|x\|_X u)=\|x\|_X Tu=0_Y. \end{align*} Since $x\in X$ was arbitrary, $T=0_{X,Y}$.[/step]

[guided]We must prove the exact definiteness axiom: the operator norm is zero precisely for the zero operator. First consider the zero operator. Let \begin{align*} 0_{X,Y}:X&\to Y \end{align*} be the map defined by $0_{X,Y}x=0_Y$ for every $x\in X$. For every $x\in X$ satisfying $\|x\|_X\le 1$, the norm of the image is \begin{align*} \|0_{X,Y}x\|_Y=\|0_Y\|_Y=0. \end{align*} Therefore the set whose supremum defines $\|0_{X,Y}\|_{\mathcal L(X,Y)}$ contains only the number $0$, and hence \begin{align*} \|0_{X,Y}\|_{\mathcal L(X,Y)}=0. \end{align*} Now assume $T\in\mathcal L(X,Y)$ and \begin{align*} \|T\|_{\mathcal L(X,Y)}=0. \end{align*} We want to prove that $T$ sends every vector in $X$ to $0_Y$. Let $x\in X$ be arbitrary. If $x=0_X$, linearity gives $Tx=T0_X=0_Y$. It remains to handle the case $x\ne 0_X$. The operator norm only controls vectors in the unit ball, so we scale $x$ into that unit ball. Define \begin{align*} u:=\frac{x}{\|x\|_X}\in X. \end{align*} Because $x\ne 0_X$, the norm axiom in $X$ gives $\|x\|_X>0$, so this definition is valid. By homogeneity of the norm on $X$, \begin{align*} \|u\|_X=\left\|\frac{x}{\|x\|_X}\right\|_X=1. \end{align*} Thus $u$ is one of the vectors appearing in the supremum defining the operator norm. Consequently, \begin{align*} 0\le \|Tu\|_Y\le \|T\|_{\mathcal L(X,Y)}=0. \end{align*} So $\|Tu\|_Y=0$, and definiteness of the norm on $Y$ implies $Tu=0_Y$. Since $x=\|x\|_X u$, linearity of $T$ gives \begin{align*} Tx=T(\|x\|_X u)=\|x\|_X Tu=0_Y. \end{align*} Thus every $x\in X$ satisfies $Tx=0_Y$, so $T=0_{X,Y}$.[/guided]

[step:Prove absolute homogeneity by pulling scalars through the supremum] Let $\alpha\in\mathbb F$ and $T\in\mathcal L(X,Y)$. If $\alpha=0$, then $\alpha T=0_{X,Y}$, so definiteness for the zero operator gives \begin{align*} \|\alpha T\|_{\mathcal L(X,Y)}=0=|\alpha|\,\|T\|_{\mathcal L(X,Y)}. \end{align*} Assume now that $\alpha\ne 0$. For every $x\in X$ with $\|x\|_X\le 1$, homogeneity of the norm on $Y$ gives \begin{align*} \|(\alpha T)x\|_Y=\|\alpha Tx\|_Y=|\alpha|\,\|Tx\|_Y. \end{align*} Since $|\alpha|>0$, multiplication by $|\alpha|$ preserves order and satisfies $\sup(|\alpha|A)=|\alpha|\sup A$ for every nonempty bounded subset $A\subset[0,\infty)$. Taking the supremum over all $x\in X$ with $\|x\|_X\le 1$ therefore yields \begin{align*} \|\alpha T\|_{\mathcal L(X,Y)}=|\alpha|\,\|T\|_{\mathcal L(X,Y)}. \end{align*} Thus the operator norm is absolutely homogeneous. [/step]

[step:Prove the triangle inequality by estimating on the unit ball] Let $S,T\in\mathcal L(X,Y)$. For every $x\in X$ with $\|x\|_X\le 1$, the triangle inequality in $Y$ gives \begin{align*} \|(S+T)x\|_Y=\|Sx+Tx\|_Y\le \|Sx\|_Y+\|Tx\|_Y. \end{align*} By the definition of the operator norm, since $\|x\|_X\le 1$, \begin{align*} \|Sx\|_Y\le \|S\|_{\mathcal L(X,Y)} \end{align*} and \begin{align*} \|Tx\|_Y\le \|T\|_{\mathcal L(X,Y)}. \end{align*} Therefore \begin{align*} \|(S+T)x\|_Y\le \|S\|_{\mathcal L(X,Y)}+\|T\|_{\mathcal L(X,Y)}. \end{align*} Taking the supremum over all $x\in X$ with $\|x\|_X\le 1$ gives \begin{align*} \|S+T\|_{\mathcal L(X,Y)}\le \|S\|_{\mathcal L(X,Y)}+\|T\|_{\mathcal L(X,Y)}. \end{align*} [/step]

[step:Conclude that all norm axioms hold] The preceding steps show that the map $T\mapsto \|T\|_{\mathcal L(X,Y)}$ is nonnegative and finite-valued on $\mathcal L(X,Y)$, vanishes only at the zero operator, is absolutely homogeneous over the scalar field $\mathbb F$, and satisfies the triangle inequality. Hence it is a norm on the [vector space](/page/Vector%20Space) $\mathcal L(X,Y)$. [/step]