[proofplan] We first make the bounded-function structure explicit: pointwise sums and scalar multiples of bounded functions are bounded because the norm on $V$ satisfies the triangle inequality and absolute homogeneity. The [vector space](/page/Vector%20Space) axioms then follow by evaluating every identity at an arbitrary point of $S$ and using the corresponding axiom in $V$. Finally, we verify the norm axioms for $\|\cdot\|_\infty$ directly from the order properties of the supremum and the norm axioms in $V$; the nonemptiness of $S$ is used in the definiteness step. [/proofplan]

[step:Show pointwise operations preserve boundedness]Let $f:S\to V$ and $g:S\to V$ be bounded functions. Choose [real numbers](/page/Real%20Numbers) $M_f\geq 0$ and $M_g\geq 0$ such that $\|f(s)\|_V\leq M_f$ and $\|g(s)\|_V\leq M_g$ for every $s\in S$. Define the pointwise sum $f+g:S\to V$ by \begin{align*} (f+g)(s) := f(s)+g(s). \end{align*} For every $s\in S$, the triangle inequality in $V$ gives \begin{align*} \|(f+g)(s)\|_V = \|f(s)+g(s)\|_V \leq \|f(s)\|_V+\|g(s)\|_V \leq M_f+M_g. \end{align*} Thus $f+g$ is bounded, so $f+g\in B(S,V)$. Let $\lambda\in\mathbb K$. Define the pointwise scalar multiple $\lambda f:S\to V$ by \begin{align*} (\lambda f)(s) := \lambda f(s). \end{align*} For every $s\in S$, absolute homogeneity of the norm on $V$ gives \begin{align*} \|(\lambda f)(s)\|_V = \|\lambda f(s)\|_V = |\lambda|\,\|f(s)\|_V \leq |\lambda|M_f. \end{align*} Hence $\lambda f$ is bounded, so $\lambda f\in B(S,V)$.[/step]

[guided]The first issue is closure: before calling $B(S,V)$ a vector space, we must check that the proposed pointwise operations stay inside $B(S,V)$. Let $f:S\to V$ and $g:S\to V$ be bounded functions. By boundedness, there are real numbers $M_f\geq 0$ and $M_g\geq 0$ such that \begin{align*} \|f(s)\|_V\leq M_f \end{align*} for every $s\in S$, and \begin{align*} \|g(s)\|_V\leq M_g \end{align*} for every $s\in S$. Define $f+g:S\to V$ pointwise by \begin{align*} (f+g)(s):=f(s)+g(s). \end{align*} This is a function from $S$ to $V$ because $f(s),g(s)\in V$ and $V$ is closed under vector addition. For each $s\in S$, the triangle inequality in the [normed vector space](/page/Normed%20Vector%20Space) $V$ gives \begin{align*} \|(f+g)(s)\|_V=\|f(s)+g(s)\|_V\leq \|f(s)\|_V+\|g(s)\|_V. \end{align*} Using the chosen bounds for $f$ and $g$, we obtain \begin{align*} \|(f+g)(s)\|_V\leq M_f+M_g. \end{align*} The number $M_f+M_g$ is independent of $s$, so $f+g$ is bounded. Therefore $f+g\in B(S,V)$. Now let $\lambda\in\mathbb K$. Define $\lambda f:S\to V$ pointwise by \begin{align*} (\lambda f)(s):=\lambda f(s). \end{align*} This is a function from $S$ to $V$ because scalar multiplication in $V$ sends $\mathbb K\times V$ to $V$. For every $s\in S$, absolute homogeneity of the norm on $V$ gives \begin{align*} \|(\lambda f)(s)\|_V=\|\lambda f(s)\|_V=|\lambda|\,\|f(s)\|_V. \end{align*} Since $\|f(s)\|_V\leq M_f$, this implies \begin{align*} \|(\lambda f)(s)\|_V\leq |\lambda|M_f. \end{align*} The bound $|\lambda|M_f$ is finite and independent of $s$, so $\lambda f$ is bounded. Therefore $\lambda f\in B(S,V)$.[/guided]

[step:Verify the vector space axioms pointwise] Let $0_V\in V$ denote the zero vector, and define the zero function $0_{B}:S\to V$ by \begin{align*} 0_B(s):=0_V. \end{align*} The function $0_B$ is bounded because $\|0_B(s)\|_V=0$ for every $s\in S$. If $f\in B(S,V)$, define $-f:S\to V$ by \begin{align*} (-f)(s):=-f(s). \end{align*} This is bounded by the scalar multiplication closure already proved, with scalar $-1\in\mathbb K$. It remains only to verify the algebraic identities. Let $f,g,h\in B(S,V)$ and let $\lambda,\mu\in\mathbb K$. For every $s\in S$, associativity and commutativity of addition, the zero and inverse laws, associativity of scalar multiplication, and the distributive laws are exactly the corresponding identities in $V$: \begin{align*} ((f+g)+h)(s)=f(s)+(g(s)+h(s))=(f+(g+h))(s). \end{align*} \begin{align*} (f+g)(s)=f(s)+g(s)=g(s)+f(s)=(g+f)(s). \end{align*} \begin{align*} (f+0_B)(s)=f(s)+0_V=f(s). \end{align*} \begin{align*} (f+(-f))(s)=f(s)-f(s)=0_V=0_B(s). \end{align*} \begin{align*} ((\lambda\mu)f)(s)=(\lambda\mu)f(s)=\lambda(\mu f(s))=(\lambda(\mu f))(s). \end{align*} \begin{align*} ((\lambda+\mu)f)(s)=(\lambda+\mu)f(s)=\lambda f(s)+\mu f(s)=(\lambda f+\mu f)(s). \end{align*} \begin{align*} (\lambda(f+g))(s)=\lambda(f(s)+g(s))=\lambda f(s)+\lambda g(s)=(\lambda f+\lambda g)(s). \end{align*} \begin{align*} (1f)(s)=1f(s)=f(s). \end{align*} Since two functions $S\to V$ are equal exactly when they agree at every point of $S$, all vector space axioms hold in $B(S,V)$. [/step]

[step:Show the supremum norm is finite and nonnegative] Let $f\in B(S,V)$. By boundedness, there exists $M\geq 0$ such that $\|f(s)\|_V\leq M$ for every $s\in S$. The set \begin{align*} A_f:=\{\|f(s)\|_V:s\in S\}\subset \mathbb R \end{align*} is nonempty because $S$ is nonempty, is bounded above by $M$, and is contained in $[0,\infty)$ because $\|\cdot\|_V$ is nonnegative. Hence $\sup A_f$ exists in $\mathbb R$, is finite, and satisfies $\sup A_f\geq 0$. Therefore $\|f\|_\infty$ is a well-defined nonnegative real number. [/step]

[step:Prove absolute homogeneity of the supremum norm] Let $f\in B(S,V)$ and $\lambda\in\mathbb K$. For every $s\in S$, \begin{align*} \|(\lambda f)(s)\|_V=|\lambda|\,\|f(s)\|_V. \end{align*} If $\lambda=0$, then $\lambda f=0_B$, so $\|\lambda f\|_\infty=0=|\lambda|\,\|f\|_\infty$. If $\lambda\neq 0$, multiplication by the positive real number $|\lambda|$ preserves suprema of bounded nonempty subsets of $\mathbb R$. Therefore \begin{align*} \|\lambda f\|_\infty=\sup_{s\in S}|\lambda|\,\|f(s)\|_V=|\lambda|\sup_{s\in S}\|f(s)\|_V=|\lambda|\,\|f\|_\infty. \end{align*} Thus $\|\cdot\|_\infty$ is absolutely homogeneous. [/step]

[step:Prove the triangle inequality for the supremum norm] Let $f,g\in B(S,V)$. For every $s\in S$, the triangle inequality in $V$ gives \begin{align*} \|(f+g)(s)\|_V\leq \|f(s)\|_V+\|g(s)\|_V. \end{align*} Since $\|f(s)\|_V\leq \|f\|_\infty$ and $\|g(s)\|_V\leq \|g\|_\infty$ for every $s\in S$, we obtain \begin{align*} \|(f+g)(s)\|_V\leq \|f\|_\infty+\|g\|_\infty. \end{align*} Thus $\|f\|_\infty+\|g\|_\infty$ is an upper bound for the nonempty set $\{\|(f+g)(s)\|_V:s\in S\}$. Taking the supremum over $s\in S$ gives \begin{align*} \|f+g\|_\infty\leq \|f\|_\infty+\|g\|_\infty. \end{align*} [/step]

[step:Prove definiteness using the nonemptiness of $S$] The zero function satisfies $\|0_B\|_\infty=0$ because $\|0_B(s)\|_V=0$ for every $s\in S$. Conversely, suppose $f\in B(S,V)$ and $\|f\|_\infty=0$. For every $s\in S$, the definition of supremum gives \begin{align*} 0\leq \|f(s)\|_V\leq \sup_{t\in S}\|f(t)\|_V=\|f\|_\infty=0. \end{align*} Hence $\|f(s)\|_V=0$ for every $s\in S$. By definiteness of the norm on $V$, this implies $f(s)=0_V$ for every $s\in S$. Therefore $f=0_B$. We have shown that $B(S,V)$ is a vector space over $\mathbb K$ and that $\|\cdot\|_\infty$ is finite-valued, nonnegative, absolutely homogeneous, satisfies the triangle inequality, and is definite. Hence $\|\cdot\|_\infty$ is a norm on $B(S,V)$. [/step]