[proofplan] We first prove that the pointwise operations preserve boundedness by writing explicit numerical bounds for $f+g$, $\lambda f$, and $fg$. We then verify the algebraic axioms pointwise, reducing each identity in $B(X)$ to the corresponding identity in $\mathbb R$. Finally, we prove that the supremum expression is a norm and that it is submultiplicative, which gives the normed algebra structure. [/proofplan] [step:Show that the pointwise operations preserve boundedness] Define the zero function $0_X:X\to\mathbb R$ by $0_X(x)=0$ for every $x\in X$. Since $|0_X(x)|=0$ for every $x\in X$, we have $0_X\in B(X)$. Let $f,g\in B(X)$ and let $\lambda\in\mathbb R$. By the definition of $B(X)$, choose $M_f,M_g\in[0,\infty)$ such that \begin{align*} |f(x)|\le M_f \text{ and } |g(x)|\le M_g \text{ for every } x\in X. \end{align*} For every $x\in X$, the triangle inequality in $\mathbb R$ gives \begin{align*} |(f+g)(x)|=|f(x)+g(x)|\le |f(x)|+|g(x)|\le M_f+M_g. \end{align*} Thus $f+g\in B(X)$. For every $x\in X$, multiplicativity of absolute value gives \begin{align*} |(\lambda f)(x)|=|\lambda f(x)|=|\lambda|\,|f(x)|\le |\lambda|M_f. \end{align*} Since $|\lambda|M_f\in[0,\infty)$, it follows that $\lambda f\in B(X)$. For every $x\in X$, \begin{align*} |(fg)(x)|=|f(x)g(x)|=|f(x)|\,|g(x)|\le M_fM_g. \end{align*} Since $M_fM_g\in[0,\infty)$, we have $fg\in B(X)$. [guided] We need to check that the pointwise formulas do not take us outside $B(X)$. The definition of $B(X)$ asks for one finite nonnegative number that bounds the absolute value of the function at every point of $X$. First define the zero function $0_X:X\to\mathbb R$ by $0_X(x)=0$ for every $x\in X$. The bound $0\in[0,\infty)$ satisfies $|0_X(x)|=0\le 0$ for every $x\in X$, so $0_X\in B(X)$. Now let $f,g\in B(X)$ and let $\lambda\in\mathbb R$. Since $f$ and $g$ are bounded, there exist numbers $M_f,M_g\in[0,\infty)$ such that \begin{align*} |f(x)|\le M_f \text{ and } |g(x)|\le M_g \text{ for every } x\in X. \end{align*} For addition, the pointwise operation gives $(f+g)(x)=f(x)+g(x)$. Applying the triangle inequality in $\mathbb R$ at each point $x\in X$ yields \begin{align*} |(f+g)(x)|=|f(x)+g(x)|\le |f(x)|+|g(x)|\le M_f+M_g. \end{align*} The number $M_f+M_g$ is finite and nonnegative, so it is a valid global bound. Hence $f+g\in B(X)$. For scalar multiplication, the pointwise operation gives $(\lambda f)(x)=\lambda f(x)$. Multiplicativity of absolute value gives \begin{align*} |(\lambda f)(x)|=|\lambda f(x)|=|\lambda|\,|f(x)|\le |\lambda|M_f. \end{align*} The number $|\lambda|M_f$ is finite and nonnegative, so $\lambda f\in B(X)$. For multiplication, the pointwise operation gives $(fg)(x)=f(x)g(x)$. Again using multiplicativity of absolute value, \begin{align*} |(fg)(x)|=|f(x)g(x)|=|f(x)|\,|g(x)|\le M_fM_g. \end{align*} The number $M_fM_g$ is finite and nonnegative, so $fg\in B(X)$. This proves closure under all three pointwise operations. [/guided] [/step] [step:Verify the vector space and algebra identities pointwise] Let $f,g,h\in B(X)$ and let $\lambda,\mu\in\mathbb R$. The closure just proved ensures that all functions appearing below belong to $B(X)$. For every $x\in X$, associativity and commutativity of addition in $\mathbb R$ give \begin{align*} ((f+g)+h)(x)=(f+(g+h))(x) \end{align*} and \begin{align*} (f+g)(x)=(g+f)(x). \end{align*} The function $0_X:X\to\mathbb R$ is an additive identity because $(f+0_X)(x)=f(x)$ for every $x\in X$. The function $-f:X\to\mathbb R$ defined by $(-f)(x)=-f(x)$ is the scalar multiple $(-1)f$, hence belongs to $B(X)$, and satisfies $(f+(-f))(x)=0$ for every $x\in X$. For scalar multiplication, the real-number identities give, for every $x\in X$, \begin{align*} (\lambda(f+g))(x)=((\lambda f)+(\lambda g))(x), \end{align*} \begin{align*} ((\lambda+\mu)f)(x)=(\lambda f+\mu f)(x), \end{align*} \begin{align*} ((\lambda\mu)f)(x)=(\lambda(\mu f))(x), \end{align*} and \begin{align*} (1f)(x)=f(x). \end{align*} Thus $B(X)$ is a real [vector space](/page/Vector%20Space) under pointwise addition and scalar multiplication. For multiplication, associativity and commutativity of multiplication in $\mathbb R$ give, for every $x\in X$, \begin{align*} ((fg)h)(x)=(f(gh))(x) \end{align*} and \begin{align*} (fg)(x)=(gf)(x). \end{align*} The distributive laws in $\mathbb R$ give \begin{align*} (f(g+h))(x)=(fg+fh)(x) \end{align*} and \begin{align*} ((f+g)h)(x)=(fh+gh)(x). \end{align*} The constant-one function $1_X:X\to\mathbb R$ defined by $1_X(x)=1$ for every $x\in X$ belongs to $B(X)$ because $|1_X(x)|\le 1$ for every $x\in X$, and it satisfies $(1_Xf)(x)=f(x)=(f1_X)(x)$ for every $x\in X$. Therefore $B(X)$ is a unital commutative algebra over $\mathbb R$. [/step] [step:Prove that the supremum formula defines a norm] Let $f\in B(X)$. Since $X$ is nonempty, the set \begin{align*} A_f:=\{|f(x)|:x\in X\} \end{align*} is a nonempty subset of $[0,\infty)$. Since $f\in B(X)$, there exists $M\in[0,\infty)$ such that $|f(x)|\le M$ for every $x\in X$, so $A_f$ is bounded above. Hence $\|f\|_\infty=\sup A_f$ is a finite real number and $\|f\|_\infty\ge 0$. If $f=0_X$, then $A_f=\{0\}$, so $\|f\|_\infty=0$. Conversely, if $\|f\|_\infty=0$, then for every $x\in X$ the number $|f(x)|$ belongs to $A_f$ and hence satisfies $|f(x)|\le \sup A_f=0$. Since also $|f(x)|\ge 0$, we get $|f(x)|=0$, so $f(x)=0$. Thus $f=0_X$. Let $\lambda\in\mathbb R$. For every $x\in X$, \begin{align*} |(\lambda f)(x)|=|\lambda|\,|f(x)|. \end{align*} If $\lambda=0$, then $\lambda f=0_X$ and $\|\lambda f\|_\infty=0=|\lambda|\,\|f\|_\infty$. If $\lambda\ne 0$, multiplying the set $A_f$ by the positive number $|\lambda|$ gives \begin{align*} \|\lambda f\|_\infty=\sup_{x\in X}|\lambda|\,|f(x)|=|\lambda|\sup_{x\in X}|f(x)|=|\lambda|\,\|f\|_\infty. \end{align*} Let $f,g\in B(X)$. For every $x\in X$, \begin{align*} |(f+g)(x)|\le |f(x)|+|g(x)|\le \|f\|_\infty+\|g\|_\infty. \end{align*} Thus $\|f\|_\infty+\|g\|_\infty$ is an upper bound for the set $\{|(f+g)(x)|:x\in X\}$. Taking the least upper bound gives \begin{align*} \|f+g\|_\infty\le \|f\|_\infty+\|g\|_\infty. \end{align*} Therefore $\|\cdot\|_\infty$ is a norm on the real vector space $B(X)$. [/step] [step:Prove the algebra norm inequality] Let $f,g\in B(X)$. Since $\|f\|_\infty$ and $\|g\|_\infty$ are upper bounds for the absolute values of $f$ and $g$, respectively, for every $x\in X$ we have \begin{align*} |f(x)|\le \|f\|_\infty \end{align*} and \begin{align*} |g(x)|\le \|g\|_\infty. \end{align*} Therefore, for every $x\in X$, \begin{align*} |(fg)(x)|=|f(x)g(x)|=|f(x)|\,|g(x)|\le \|f\|_\infty\|g\|_\infty. \end{align*} Thus $\|f\|_\infty\|g\|_\infty$ is an upper bound for $\{|(fg)(x)|:x\in X\}$. Taking the supremum over $x\in X$ gives \begin{align*} \|fg\|_\infty\le \|f\|_\infty\|g\|_\infty. \end{align*} Combining the pointwise algebra structure, the norm axioms, and this submultiplicative estimate, $B(X)$ is a unital normed algebra over $\mathbb R$. [/step]