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Proof

(related to Proposition: Calculations with Uniformly Convergent Functions)

By hypothesis, $\mathbb F$ is either the field of real numbers or the field of complex numbers, let $D\subset \mathbb F.$ Let $f_n,g_n,f,g:D\to\mathbb F$ be functions, $\alpha_n,\alpha\in\mathbb F,$ and let $f_n\to f,$ $g_n\to g$ are uniformly convergent, and $(\alpha_n)_{n\in\mathbb N}$ be a sequence with the limit $\alpha_n\to \alpha.$

• Ad $f_n+g_n\to f+g$
  • Using the supremum norm, and applying the triangle inequality, we get $||(f_n+g_n)-(f+g)||_\infty\le ||f_n-f||_\infty +||g_n-g||_\infty.$
  • By assumption, $f_n\to f$ and $g_n\to g$.
  • Therefore, using the criterion supremum norm and uniform convergence, from the assumption and the above inequality it follows $$\lim_{n\to\infty}||(f_n+g_n)-(f+g)||_\infty=0,$$ in other words $f_n+g_n$ is uniformly convergent to $f+g$ on $D.$
• Ad $\alpha f_n\to \alpha f$
  • Similarly, $||\alpha f_n-\alpha f||_\infty\le |\alpha|\cdot ||f_n-f||_\infty.$
  • Again, using the criterion supremum norm and uniform convergence, from $f_n\to f$ and the above inequality oit follows $$\lim_{n\to\infty}||(\alpha f_n-\alpha f)||_\infty=0.$$
  • Thus $\alpha f_n\to \alpha f$ on $D.$

Now, assume that $f_n,g_n$ are bounded and that the sequence $\alpha_n$ is convergent to $\alpha.$

• We will first show that $f,g$ are bounded.
  • By assumption $f_n,g_n$ are bounded for all $n\in\mathbb N.$
  • By definition of bounded, there is a positive real number $B > 0$ such that $|f_n(x)| < B,$ $|g_n(x)| < B$ for all $n\in\mathbb N$ and all $x\in D.$
  • By definition of supremum norms, this yieds * $||f||_\infty < B$ and * $||g||_\infty < B.$
  • Define $\phi_n,\psi:D\to\mathbb N$ by $f_n=\phi_n+f$ and $g_n=\psi_n+g$ for all $n\in\mathbb N.$
  • From the triangle inequality, it follows for the supremum norms $||f_n||_\infty\le ||\phi_n||_\infty + ||f||_\infty$ and $||g_n||_\infty\le ||\psi_n||_\infty + ||g||_\infty.$
  • Note that since $f_n\to f$ and $g_n\to g$ uniformly, we have $\phi_n\to 0$ and $\psi\to 0$ uniformly.
  • Therefore, * $||f_n||_\infty\le 1 + ||f||_\infty$ and * $||g_n||_\infty\le 1 + ||g||_\infty.$
  • On the other hand, from the uniform convergence and the reverse triangle inequality it follows for $\epsilon = 1$ there is an index $N\in\mathbb N$ such that for all $x\in D$ and $n\ge N$ * $|f(x)|-|f_n(x)|\le \left||f(x)|-|f_n(x)|\right|\le |f(x)-f_n(x)| < 1$ and * $|g(x)|-|g_n(x)|\le \left||g(x)|-|g_n(x)|\right|\le |g(x)-g_n(x)| < 1.$
  • This yields for all $x\in D$ and $n\ge N$ * $|f(x)| < 1 + |f_n(x)|\le 1+||f_n||_\infty < 1+ B$ and * $|g(x)| < 1 + |g_n(x)|\le 1+||g_n||_\infty < 1+ B$.
  • It follows that $f,g$ are bounded on $D.$
• Ad $f_ng_n\to fg$
  • Note that $f_ng_n-fg=(f_n-f)g_n+(g_n-g)f.$
  • Therefore, $$\begin{array}{rcl}||f_ng_n-fg||_\infty&=&||(f_n-f)g_n+(g_n-g)f||_\infty\\ &\le&||(f_n-f)g_n||_\infty+||(g_n-g)f||_\infty\\ &\le&||(f_n-f)||_\infty||g_n||_\infty+||(g_n-g)||_\infty||f||_\infty\\ &\le&||(f_n-f)||_\infty B+||(g_n-g)||_\infty B.\\ \end{array}$$
  • since $f_n\to f$ and $g_n\to g$ uniformly, we have $f_ng_n\to fg$ uniformly.
• Ad $\alpha_n f_n\to \alpha f$
  • This follows from $f_ng_n\to fg$ as a special case for the constant functions $g_n(x)=:\alpha_n$ for all $x\in D.$
    ∎

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