[proofplan] We first show that pointwise addition and scalar multiplication keep bounded linear maps inside $\mathcal{L}(V,W)$. Then the [vector space](/page/Vector%20Space) axioms follow by evaluating each identity at an arbitrary vector in $V$ and using the corresponding vector space axiom in $W$. Finally, we verify the norm axioms for the [operator norm](/page/Operator%20Norm) directly from the supremum definition over the closed unit ball of $V$. [/proofplan]

[step:Show that pointwise operations preserve bounded linearity]Let $S,T\in\mathcal{L}(V,W)$ and let $\alpha\in\mathbb{F}$. Define the maps $S+T:V\to W$ and $\alpha T:V\to W$ by \begin{align*} (S+T)(v)=S(v)+T(v) \end{align*} and \begin{align*} (\alpha T)(v)=\alpha T(v) \end{align*} for $v\in V$. For all $v_1,v_2\in V$ and all $\beta,\gamma\in\mathbb{F}$, linearity of $S$ and $T$ gives \begin{align*} (S+T)(\beta v_1+\gamma v_2)=\beta(S+T)(v_1)+\gamma(S+T)(v_2). \end{align*} Thus $S+T$ is $\mathbb{F}$-linear. Also, \begin{align*} (\alpha T)(\beta v_1+\gamma v_2)=\beta(\alpha T)(v_1)+\gamma(\alpha T)(v_2), \end{align*} so $\alpha T$ is $\mathbb{F}$-linear. Since $S$ and $T$ are bounded, there exist constants $C_S,C_T\in[0,\infty)$ such that \begin{align*} \|S(v)\|_W\leq C_S\|v\|_V \end{align*} and \begin{align*} \|T(v)\|_W\leq C_T\|v\|_V \end{align*} for every $v\in V$. By the triangle inequality in $W$, \begin{align*} \|(S+T)(v)\|_W\leq (C_S+C_T)\|v\|_V. \end{align*} Hence $S+T$ is bounded. Similarly, \begin{align*} \|(\alpha T)(v)\|_W=|\alpha|\,\|T(v)\|_W\leq |\alpha|C_T\|v\|_V, \end{align*} so $\alpha T$ is bounded. Therefore $S+T\in\mathcal{L}(V,W)$ and $\alpha T\in\mathcal{L}(V,W)$.[/step]

[guided]We need to prove that the operations stated in the theorem actually stay inside the set $\mathcal{L}(V,W)$. An element of $\mathcal{L}(V,W)$ is not just any map $V\to W$: it must be $\mathbb{F}$-linear and bounded. Take $S,T\in\mathcal{L}(V,W)$ and $\alpha\in\mathbb{F}$. Define $S+T:V\to W$ and $\alpha T:V\to W$ pointwise by \begin{align*} (S+T)(v)=S(v)+T(v) \end{align*} and \begin{align*} (\alpha T)(v)=\alpha T(v). \end{align*} First we check linearity. If $v_1,v_2\in V$ and $\beta,\gamma\in\mathbb{F}$, then linearity of $S$ and $T$ gives \begin{align*} (S+T)(\beta v_1+\gamma v_2)=S(\beta v_1+\gamma v_2)+T(\beta v_1+\gamma v_2). \end{align*} Using linearity of each operator separately, this becomes \begin{align*} (S+T)(\beta v_1+\gamma v_2)=\beta S(v_1)+\gamma S(v_2)+\beta T(v_1)+\gamma T(v_2). \end{align*} Regrouping in the vector space $W$, we get \begin{align*} (S+T)(\beta v_1+\gamma v_2)=\beta(S+T)(v_1)+\gamma(S+T)(v_2). \end{align*} Thus $S+T$ is $\mathbb{F}$-linear. For scalar multiplication, linearity of $T$ gives \begin{align*} (\alpha T)(\beta v_1+\gamma v_2)=\alpha T(\beta v_1+\gamma v_2). \end{align*} Therefore \begin{align*} (\alpha T)(\beta v_1+\gamma v_2)=\alpha\beta T(v_1)+\alpha\gamma T(v_2). \end{align*} Since scalar multiplication in $W$ is associative and commutative at the scalar level, this is \begin{align*} (\alpha T)(\beta v_1+\gamma v_2)=\beta(\alpha T)(v_1)+\gamma(\alpha T)(v_2). \end{align*} Thus $\alpha T$ is also $\mathbb{F}$-linear. Now we check boundedness. Since $S$ and $T$ are bounded, there exist constants $C_S,C_T\in[0,\infty)$ such that, for every $v\in V$, \begin{align*} \|S(v)\|_W\leq C_S\|v\|_V \end{align*} and \begin{align*} \|T(v)\|_W\leq C_T\|v\|_V. \end{align*} Using the triangle inequality in the [normed space](/page/Normed%20Space) $W$, we obtain \begin{align*} \|(S+T)(v)\|_W=\|S(v)+T(v)\|_W\leq \|S(v)\|_W+\|T(v)\|_W. \end{align*} Substituting the two boundedness estimates gives \begin{align*} \|(S+T)(v)\|_W\leq (C_S+C_T)\|v\|_V. \end{align*} The constant $C_S+C_T$ is finite, so $S+T$ is bounded. For $\alpha T$, homogeneity of the norm on $W$ gives \begin{align*} \|(\alpha T)(v)\|_W=\|\alpha T(v)\|_W=|\alpha|\,\|T(v)\|_W. \end{align*} Using the boundedness estimate for $T$, we get \begin{align*} \|(\alpha T)(v)\|_W\leq |\alpha|C_T\|v\|_V. \end{align*} Thus $\alpha T$ is bounded. We have proved both closure properties: $S+T\in\mathcal{L}(V,W)$ and $\alpha T\in\mathcal{L}(V,W)$.[/guided]

[step:Verify the vector space axioms pointwise] Define the zero operator $0_{\mathcal{L}}:V\to W$ by \begin{align*} 0_{\mathcal{L}}(v)=0_W \end{align*} for every $v\in V$, where $0_W$ is the zero vector of $W$. Let $0_V$ denote the zero vector of $V$. The map $0_{\mathcal{L}}$ is linear and bounded because \begin{align*} \|0_{\mathcal{L}}(v)\|_W=0\leq 0\|v\|_V \end{align*} for every $v\in V$. Let $R,S,T\in\mathcal{L}(V,W)$ and let $\alpha,\beta\in\mathbb{F}$. Each vector space identity in $\mathcal{L}(V,W)$ is checked by evaluating at an arbitrary $v\in V$. For example, \begin{align*} ((R+S)+T)(v)=(R(v)+S(v))+T(v)=R(v)+(S(v)+T(v))=(R+(S+T))(v), \end{align*} so addition is associative. Evaluating at an arbitrary $v\in V$ gives $(R+S)(v)=R(v)+S(v)=S(v)+R(v)=(S+R)(v)$, so addition is commutative. Likewise $(R+0_{\mathcal{L}})(v)=R(v)+0_W=R(v)$, so $0_{\mathcal{L}}$ is an additive identity. Also $(R+(-R))(v)=R(v)+(-R(v))=0_W$, so additive inverses exist. For $\alpha,\beta\in\mathbb{F}$, we have $(1R)(v)=1\cdot R(v)=R(v)$, $((\alpha+\beta)R)(v)=(\alpha+\beta)R(v)=\alpha R(v)+\beta R(v)=((\alpha R)+(\beta R))(v)$, $(\alpha(R+S))(v)=\alpha(R(v)+S(v))=((\alpha R)+(\alpha S))(v)$, and $((\alpha\beta)R)(v)=\alpha(\beta R(v))=(\alpha(\beta R))(v)$. These identities verify scalar identity, distributivity, and scalar associativity pointwise. Hence $\mathcal{L}(V,W)$ is a vector space over $\mathbb{F}$. [/step]

[step:Prove nonnegativity and definiteness of the operator norm] Let $T\in\mathcal{L}(V,W)$. Since $\|T(v)\|_W\geq 0$ for every $v\in V$, the supremum defining $\|T\|_{\mathcal{L}(V,W)}$ is nonnegative. Assume $\|T\|_{\mathcal{L}(V,W)}=0$. Let $v\in V$. If $v=0_V$, then $T(v)=0_W$ by linearity. If $v\neq 0_V$, define $u\in V$ by \begin{align*} u=\frac{1}{\|v\|_V}v. \end{align*} Then $\|u\|_V=1$, so \begin{align*} 0\leq \|T(u)\|_W\leq \|T\|_{\mathcal{L}(V,W)}=0. \end{align*} Thus $\|T(u)\|_W=0$, and the definiteness axiom for the norm on $W$ gives $T(u)=0_W$. By linearity, \begin{align*} T(v)=\|v\|_V T(u)=0_W. \end{align*} Since $v\in V$ was arbitrary, $T=0_{\mathcal{L}}$. Conversely, if $T=0_{\mathcal{L}}$, then $\|T(v)\|_W=0$ for all $v\in V$, so \begin{align*} \|T\|_{\mathcal{L}(V,W)}=0. \end{align*} Therefore $\|T\|_{\mathcal{L}(V,W)}=0$ if and only if $T=0_{\mathcal{L}}$. [/step]

[step:Prove homogeneity of the operator norm] Let $T\in\mathcal{L}(V,W)$ and let $\alpha\in\mathbb{F}$. If $\alpha=0$, then $\alpha T=0_{\mathcal{L}}$, and the preceding step gives \begin{align*} \|\alpha T\|_{\mathcal{L}(V,W)}=0=|\alpha|\,\|T\|_{\mathcal{L}(V,W)}. \end{align*} If $\alpha\neq 0$, then for every $v\in V$ with $\|v\|_V\leq 1$, \begin{align*} \|(\alpha T)(v)\|_W=|\alpha|\,\|T(v)\|_W. \end{align*} Taking the supremum over all $v\in V$ with $\|v\|_V\leq 1$ gives \begin{align*} \|\alpha T\|_{\mathcal{L}(V,W)}=|\alpha|\,\|T\|_{\mathcal{L}(V,W)}. \end{align*} Thus homogeneity holds for every $\alpha\in\mathbb{F}$. [/step]

[step:Prove the triangle inequality for the operator norm] Let $S,T\in\mathcal{L}(V,W)$. For every $v\in V$ with $\|v\|_V\leq 1$, the triangle inequality in $W$ gives \begin{align*} \|(S+T)(v)\|_W=\|S(v)+T(v)\|_W\leq \|S(v)\|_W+\|T(v)\|_W. \end{align*} By the definition of the operator norm, \begin{align*} \|S(v)\|_W\leq \|S\|_{\mathcal{L}(V,W)} \end{align*} and \begin{align*} \|T(v)\|_W\leq \|T\|_{\mathcal{L}(V,W)}. \end{align*} Therefore \begin{align*} \|(S+T)(v)\|_W\leq \|S\|_{\mathcal{L}(V,W)}+\|T\|_{\mathcal{L}(V,W)}. \end{align*} Taking the supremum over all $v\in V$ with $\|v\|_V\leq 1$ yields \begin{align*} \|S+T\|_{\mathcal{L}(V,W)}\leq \|S\|_{\mathcal{L}(V,W)}+\|T\|_{\mathcal{L}(V,W)}. \end{align*} Thus the operator norm satisfies the triangle inequality. Combining nonnegativity, definiteness, homogeneity, and the triangle inequality, $\|\cdot\|_{\mathcal{L}(V,W)}$ is a norm on $\mathcal{L}(V,W)$. Together with the vector space structure already proved, this shows that $\mathcal{L}(V,W)$ is a [normed vector space](/page/Normed%20Vector%20Space). [/step]