[proofplan] We prove the three bounds pointwise with constants independent of both the index $n$ and the point $x \in E$. The sum estimate follows from the triangle inequality for the absolute value on $\mathbb R$. The product and scalar-multiple estimates follow from multiplicativity of absolute value and the assumed uniform bounds. Since the resulting constants depend only on $M$, $N$, and $a$, they give uniform boundedness of the corresponding sequences. [/proofplan] [step:Fix an arbitrary scalar, index, and point] Let $a \in \mathbb R$, let $n \in \mathbb N$, and let $x \in E$ be arbitrary. By hypothesis, \begin{align*} |f_n(x)| \le M \end{align*} and \begin{align*} |g_n(x)| \le N. \end{align*} Because $M,N \in [0,\infty)$, the quantities $M+N$, $MN$, and $|a|M$ are nonnegative constants independent of $n$ and $x$. [/step] [step:Bound the pointwise sum by the sum of the uniform bounds] The triangle inequality for the absolute value on $\mathbb R$, applied to the [real numbers](/page/Real%20Numbers) $f_n(x)$ and $g_n(x)$, gives \begin{align*} |f_n(x)+g_n(x)| \le |f_n(x)|+|g_n(x)|. \end{align*} Using the assumed bounds on the two terms on the right-hand side gives \begin{align*} |f_n(x)+g_n(x)| \le M+N. \end{align*} [guided] We want a bound for the value of the sum at the fixed point $x$ and fixed index $n$. The only available information is the separate control of $f_n(x)$ and $g_n(x)$, so the appropriate elementary tool is the triangle inequality for real absolute value. Applied to the real numbers $f_n(x)$ and $g_n(x)$, it gives \begin{align*} |f_n(x)+g_n(x)| \le |f_n(x)|+|g_n(x)|. \end{align*} The hypotheses now control each summand: \begin{align*} |f_n(x)| \le M \end{align*} and \begin{align*} |g_n(x)| \le N. \end{align*} Substituting these two inequalities into the right-hand side yields \begin{align*} |f_n(x)+g_n(x)| \le M+N. \end{align*} The important point is that $M+N$ does not depend on the chosen index $n$ or point $x$, so this is a uniform pointwise bound for the whole sequence of sums. [/guided] [/step] [step:Bound the pointwise product by the product of the uniform bounds] Multiplicativity of absolute value on $\mathbb R$ gives \begin{align*} |f_n(x)g_n(x)| = |f_n(x)|\,|g_n(x)|. \end{align*} Since $|f_n(x)| \le M$, $|g_n(x)| \le N$, and all four quantities are nonnegative, multiplication preserves the inequalities and therefore \begin{align*} |f_n(x)g_n(x)| \le MN. \end{align*} [/step] [step:Bound the pointwise scalar multiple by the scaled uniform bound] Multiplicativity of absolute value on $\mathbb R$, applied to the real numbers $a$ and $f_n(x)$, gives \begin{align*} |a f_n(x)| = |a|\,|f_n(x)|. \end{align*} Using $|f_n(x)| \le M$ and $|a| \ge 0$, multiplication by $|a|$ preserves the inequality, so \begin{align*} |a f_n(x)| \le |a|M. \end{align*} [/step] [step:Conclude uniform boundedness of the three derived sequences] Because $a \in \mathbb R$, $n \in \mathbb N$, and $x \in E$ were arbitrary, the preceding estimates hold for every $a \in \mathbb R$, every $n \in \mathbb N$, and every $x \in E$. For each fixed $a \in \mathbb R$, the sequence of functions $(a f_n)_{n=1}^{\infty}$ is bounded by the common constant $|a|M$ on $E$. The sequence $(f_n+g_n)_{n=1}^{\infty}$ is bounded by the common constant $M+N$ on $E$, and the sequence $(f_n g_n)_{n=1}^{\infty}$ is bounded by the common constant $MN$ on $E$. Hence all three derived sequences are uniformly bounded on $E$. [/step]