[proofplan] The proof is a direct unpacking of the [operator norm](/page/Operator%20Norm). The zero vector is handled separately using linearity. For a nonzero vector $x$, we normalize it to a unit vector $u=x/\|x\|_X$, apply the definition of $\|T\|_{\mathcal L(X,Y)}$ to $u$, and then rescale using linearity and homogeneity of the norm on $Y$. [/proofplan] [step:Verify the inequality at the zero vector] Let $0_X\in X$ and $0_Y\in Y$ denote the zero vectors. Since $T:X\to Y$ is linear, $T0_X=0_Y$. Therefore \begin{align*} \|T0_X\|_Y=0\le \|T\|_{\mathcal L(X,Y)}\|0_X\|_X. \end{align*} Thus the desired inequality holds for $x=0_X$. [/step] [step:Normalize a nonzero vector and apply the definition of the operator norm] Fix $x\in X$ with $x\ne 0_X$. Define \begin{align*} u:=\frac{x}{\|x\|_X}\in X. \end{align*} Since $x\ne 0_X$, the norm axiom gives $\|x\|_X>0$, so $u$ is well-defined. By homogeneity of the norm on $X$, \begin{align*} \|u\|_X=\left\|\frac{x}{\|x\|_X}\right\|_X=\frac{1}{\|x\|_X}\|x\|_X=1. \end{align*} By the definition of the operator norm on $\mathcal L(X,Y)$, \begin{align*} \|Tu\|_Y\le \|T\|_{\mathcal L(X,Y)}. \end{align*} [guided] Fix $x\in X$ with $x\ne 0_X$. The goal is to compare $\|Tx\|_Y$ with the operator norm of $T$, and the operator norm controls $T$ on vectors of norm $1$. Therefore we first convert $x$ into a unit vector. Define \begin{align*} u:=\frac{x}{\|x\|_X}\in X. \end{align*} This definition is valid because $x\ne 0_X$, so the norm axiom for $X$ gives $\|x\|_X>0$. Using homogeneity of the norm on $X$, we compute \begin{align*} \|u\|_X=\left\|\frac{x}{\|x\|_X}\right\|_X=\frac{1}{\|x\|_X}\|x\|_X=1. \end{align*} Thus $u$ is exactly the kind of vector tested by the operator norm. Since $T\in\mathcal L(X,Y)$, its operator norm is defined by taking the supremum of $\|Tv\|_Y$ over unit vectors $v\in X$. Applying that definition to the unit vector $u$, we obtain \begin{align*} \|Tu\|_Y\le \|T\|_{\mathcal L(X,Y)}. \end{align*} [/guided] [/step] [step:Rescale the normalized estimate back to the original vector] Since $x=\|x\|_X u$ and $T$ is linear, \begin{align*} Tx=T(\|x\|_X u)=\|x\|_X Tu. \end{align*} By homogeneity of the norm on $Y$, \begin{align*} \|Tx\|_Y=\|\|x\|_X Tu\|_Y=\|x\|_X\|Tu\|_Y. \end{align*} Combining this identity with $\|Tu\|_Y\le \|T\|_{\mathcal L(X,Y)}$ gives \begin{align*} \|Tx\|_Y\le \|x\|_X\|T\|_{\mathcal L(X,Y)}. \end{align*} This proves the stated inequality for every nonzero $x\in X$, and the zero-vector case was proved above. Therefore the inequality holds for every $x\in X$. [/step]