7.4 Sums and Products of Null Sequences

7.27   Theorem (Sum theorem for null sequences.)Let $f,g$ be complex null sequences and let $\alpha\in\mbox{{\bf C}}$. Then $f+g$, $f-g$, and $\alpha f$ are null sequences.

Scratchwork for $\alpha f$: I want to find $N_{\alpha f}$ so that

$$n\geq N_{\alpha f}(\varepsilon)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert\alpha f(n)\vert<\varepsilon$$

i.e.

$$n \geq N_{\alpha f}(\varepsilon) \mbox{$\hspace{1ex}\Longrigh... ...e{1ex}$} \vert f(n)\vert<{\varepsilon\over {\vert\alpha\vert}}.$$

This suggests that I take $${N_{\alpha f}(\varepsilon)=N_f\left({\varepsilon\over {\vert\alpha\vert}}\right)}$$.

Scratchwork for $f+g$: I want to find $N_{f+g}$ so that

$$n\geq N_{f+g}(\varepsilon)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert f(n)+g(n)\vert<\varepsilon.$$

Now $\vert f(n)+g(n)\vert\leq\vert f(n)\vert+\vert g(n)\vert$, and I can make $\vert f(n)\vert+\vert g(n)\vert<\varepsilon$ by making $\vert f(n)\vert<\varepsilon /2$ and $\vert g(n)\vert<\varepsilon /2$. Hence I want $N_{f+g}(\varepsilon)>N_f(\varepsilon/2)$ and $${N_{f+g}(\varepsilon)>N_g\left({\varepsilon\over 2}\right)}$$. This suggests that I take $N_{f+g}(\varepsilon)=\max\left(N_f(\varepsilon /2),N_g(\varepsilon /2)\right)$.

Proof: Let $f,g$ be null sequences, and let $\alpha\in\mbox{{\bf C}}$. Define $N_{f+g}\colon\mbox{${\mbox{{\bf R}}}^{+}$}\to\mbox{{\bf N}}$ by

$$N_{f+g}(\varepsilon)=\max\left(N_f(\varepsilon /2),N_g(\varepsilon/2)\right).$$

Then for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n\geq N_{f+g}(\varepsilon)&\mbox{$\Longrightarrow$}&n\geq N_f(... ...2} \\ &\mbox{$\Longrightarrow$}&\vert(f+g)(n)\vert<\varepsilon. \end{eqnarray*}$$

Hence, $N_{f+g}$ is a precision function for $f+g$, and $f+g$ is a null sequence.

If $\alpha=0$ then $\alpha f=\tilde 0$ is a null sequence. Suppose $\alpha\neq 0$, and define $N_{\alpha f}\colon\mbox{${\mbox{{\bf R}}}^{+}$}\to\mbox{{\bf N}}$ by

$$N_{\alpha f}(\varepsilon)=N_f\left({\varepsilon\over {\vert\alpha\vert}}\right).$$

Then for all $n\in\mbox{{\bf Z}}$,

$$\begin{eqnarray*} n\geq N_{\alpha f}&\mbox{$\Longrightarrow$}&n\geq N_f\left({\v... ...\mbox{$\Longrightarrow$}&\vert\alpha f(n)\vert\leq \varepsilon . \end{eqnarray*}$$

Hence $N_{\alpha f}$ is a precision function for $\alpha f$, and hence $\alpha f$ is a null sequence. Since $f-g=f+(-1)g$ it follows that $f-g$ is a null sequence. $\mid\!\mid\!\mid$

7.28   Exercise (Product theorem for null sequences.) A Let $f,g$ be complex null sequences. Prove that $fg$ is a null sequence.