Exercise 3.3.8

Question

Let $(X, d)$ be a metric space, and for every positive integer $n$, let $f^{(n)}$ : $X \rightarrow \mathbb{R}$ and $g^{(n)}: X \rightarrow \mathbb{R}$ be functions. Suppose that $\left(f^{(n)}\right)_{n=1}^{\infty}$ converges uniformly to another function $f: X \rightarrow \mathbb{R}$, and that $\left(g^{(n)}\right)_{n=1}^{\infty}$ converges uniformly to another function $g: X \rightarrow \mathbb{R}$. Suppose also that the functions $\left(f^{(n)}\right)_{n=1}^{\infty}$ and $\left(g^{(n)}\right)_{n=1}^{\infty}$ are uniformly bounded. Prove that the functions $f^{(n)} g^{(n)}: X \rightarrow \mathbb{R}$ converge uniformly to $f g: X \rightarrow \mathbb{R}$.

Prakash Anand · Accepted Answer

Suppose that $\left(f^{(n)} ight)_{n=1}^{\infty}$ converges uniformly to another function $f: X ightarrow \mathbb{R}$, and that $\left(g^{(n)} ight)_{n=1}^{\infty}$ converges uniformly to another function $g: X ightarrow \mathbb{R}$. That is, given $\epsilon^{\prime}>0$ there exists $N^{\prime}>0$ such that $\left|f^{(n)}(x)-f(x) ight|<\epsilon^{\prime}$ and $\left|g^{(n)}(x)-g(x) ight|<\epsilon^{\prime}$ for all $n>N^{\prime}$ and $x \in X$. Also suppose that the functions $\left(f^{(n)} ight)_{n=1}^{\infty}$ and $\left(g^{(n)} ight)_{n=1}^{\infty}$ are uniformly bounded. That is, there exists $M>0$ such that $\left|f^{(n)}(x) ight| \leq M$ and $\left|g^{(n)}(x) ight| \leq M$ for all $n \geq 1$ and $x \in X$. And hence from the previous exercise we know that $|f(x)|,|g(x)| \leq M$ for all $x \in X$. (Note that I am being lazy and using the same $M$ as above. This is not necessarily true, but we can take the max between the two $M$ 's and just call that $M$ ). Given $\epsilon>0$ take $\epsilon^{\prime}=\epsilon / M$ and choose $N=N^{\prime}$. Thus we have that $\left|f^{(n)}(x)-f(x) ight|<\epsilon^{\prime}=$ $\epsilon / M$ and $\left|g^{(n)}(x)-g(x) ight|<\epsilon^{\prime}=\epsilon / M$ for all $n>N^{\prime}$ and $x \in X$. But, $$ \begin{aligned} \left|f^{(n)} g^{(n)}(x)-f g(x) ight| &=\left|f^{(n)} g^{(n)}(x)-f g(x)+f^{(n)} g(x)-f^{(n)} g(x) ight| \ &=\left|f^{(n)}(x)\left(g^{(n)}(x)-g(x) ight)+g(x)\left(f^{(n)}(x)-f(x) ight) ight| \ & \leq\left|f^{(n)}(x) ight|\left|g^{(n)}(x)-g(x) ight|+|g(x)|\left|f^{(n)}(x)-f(x) ight| \ & \leq M \frac{\epsilon}{M}+M \frac{\epsilon}{M}=\epsilon \end{aligned} $$ Thus $f^{(n)} g^{(n)} ightarrow f g$ uniformly.

Exercise 3.3.8

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Exercise 3.3.8

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