Exercise 3.3.6 (Uniform limits preserve boundedness)

Question

Prove Proposition 3.3.6.\par 

	extbf{Proposition 3.3.6.} Let $\left(f^{(n)}ight)_{n=1}^{\infty}$ be a sequence of functions from one metric space $\left(X, d_{X}ight)$ to another $\left(Y, d_{Y}ight)$, and suppose that this sequence converges uniformly to another function $f: X ightarrow Y$. If the functions $f^{(n)}$ are bounded on $X$ for each $n$, then the limiting function $f$ is also bounded on $X$.

Prakash Anand · Accepted Answer

\begin{proof} Suppose that $\left(f^{(n)}\right)_{n=1}^{\infty}$ is a Convergent sequence, to say $f$. Then from Lemma 1\footnote{\begin{lemma}If $\left(f^{(n)}\right)_{n=1}^{\infty}$ converges uniformly to some function $f$, then the sequence $\left(\right.$ of $f^{(n)}$ 's $)$ is uniformly Cauchy.\end{lemma}\begin{proof}Since $f^{(n)} \rightarrow f$ uniformly, we know that given $\epsilon_{1}>0$ there exists $N_{1}>0$ such that $d_{Y}\left(f^{(n)}(x), f(x)\right)<\epsilon_{1}$ for all $n>N_{1}$ and $x \in X$. So we also have $d_{Y}\left(f^{(m)}(x), f(x)\right)<\epsilon_{1}$ for all $m>N_{1}$ and $x \in X$. So given $\epsilon>0$ take $\epsilon_{1}=\epsilon / 2$ and $N=N_{1}$. Then we get that $$ d_{Y}\left(f^{(n)}(x), f^{(m)}(x)\right) \leq d_{Y}\left(f^{(n)}(x), f(x)\right)+d_{Y}\left(f(x), f^{(m)}(x)\right)<\epsilon $$ for all $n, m>N$ and $x \in X$. Hence $\left(f^{(n)}\right)_{n=1}^{\infty}$ is uniformly Cauchy.\end{proof}} we have that $\left(f^{(n)}\right)_{n=1}^{\infty}$ is also uniformly Cauchy. That is, given $\epsilon>0$ there exists $N>0$ such that $$d_{Y}\left(f^{(n)}(x), f(x)\right)<\epsilon$$ for all $n>N$ and $x \in X$. And also that $$d_{Y}\left(f^{(n)}(x), f^{(m)}(x)\right)<\epsilon$$ for all $n, m>N$ and $x \in X$. Also suppose that for each $n$ that $f^{(n)}$ is bounded. That is, there is some ball $B_{\left(Y, d_{Y}\right)}\left(y_{n}, R_{n}\right)$ such that $$f^{(n)}(x) \in B_{\left(Y, d_{Y}\right)}\left(y_{n}, R_{n}\right)$$ for all $x \in X$. That is, $d_{Y}\left(f^{(n)}(x), y_{n}\right)N$ and $x \in X$. Hence there exists $N>0$ (same $N$ as above) such that $$f^{(n)}(x) \in B_{\left(Y, d_{Y}\right)}\left(y_{n}, R_{n}+\epsilon\right)$$ for all $n>N$ and $x \in X$. But, $$d_{Y}\left(f(x), y_{n}\right) \leq d_{Y}\left(f(x), f^{(n)}(x)\right)+d_{Y}\left(f^{(n)}(x), y_{n}\right)<\epsilon+R_{n}+\epsilon=R_{n}+2 \epsilon.$$ So $f(x) \in B_{\left(Y, d_{Y}\right)}\left(y_{n}, R_{n}+2 \epsilon\right)$ for all $x \in X$. Therefore $f(x)$ is bounded on $X$. \end{proof}

Exercise 3.3.6 (Uniform limits preserve boundedness)

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Exercise 3.3.6 (Uniform limits preserve boundedness)

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