Prove or disprove each statement.
If \(g(t)\to b\) as \(t\to a\) and \(f(t)\to c\) as \(t\to b\), then \(f(g(t))\to c\) as \(t\to a\).
If \(f:\mathbb R\to\mathbb R\) is continuous and \(U\) is open, then \(f(U)\) is open.
If \(f\in C^\infty(-1,1)\) and \(|f^{(n)}(x)|\le1\) for all \(n\ge1\) and \(x\in(-1,1)\), then \(f\) is real analytic.
Solution
1. Statement (a) is false.
Take \[ a=b=c=0, \] and define \[ g(t)=0 \] for every \(t\). Then \[ g(t)\to0 \] as \(t\to0\).
Define \[ f(s)= \begin{cases} 1,&s=0,\\ 0,&s\ne0. \end{cases} \] Then \[ f(s)\to0 \] as \(s\to0\), because a limit at \(0\) does not depend on the value \(f(0)\). But \[ f(g(t))=f(0)=1 \] for all \(t\), so \[ f(g(t))\not\to0. \] Thus (a) is false.
2. Statement (b) is false.
Take the continuous constant function \[ f(x)=0 \] on \(\mathbb R\) and the open set \[ U=(0,1). \] Then \[ f(U)=\{0\}, \] which is not open in \(\mathbb R\). Hence a continuous real-valued function need not be an open map.
3. Statement (c) is true.
Fix \[ x_0\in(-1,1). \] For any \(x\in(-1,1)\) and integer \(N\ge0\), Taylor’s theorem with remainder gives a point \(\xi_N\) between \(x_0\) and \(x\) such that \[ f(x) =\sum_{k=0}^{N}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k +\frac{f^{(N+1)}(\xi_N)}{(N+1)!}(x-x_0)^{N+1}. \] The hypothesis gives \[ |f^{(N+1)}(\xi_N)|\le1, \] so the remainder satisfies \[ |R_N(x)| \le\frac{|x-x_0|^{N+1}}{(N+1)!}. \] For fixed \(x\), the right-hand side tends to \(0\) as \(N\to\infty\). Therefore \[ f(x) =\sum_{k=0}^{\infty} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k \] for every \(x\in(-1,1)\). In particular, \(f\) agrees with its Taylor series in a neighborhood of every \(x_0\), so \(f\) is real analytic.
Thus the answers are \[ \boxed{\text{(a) false,\qquad (b) false,\qquad (c) true}.} \]