91Ó°ÊÓ

First, we need to check if the given function \(\|f\|\) satisfies all the properties of a norm. A function \(\|.\|\) is called a norm if it satisfies the following three properties for all \(f, g \in C^{1}([a, b])\) and all scalars \(c\): 1. \(\|f\| \geq 0\), and \(\|f\| = 0\) if and only if \(f = 0\). 2. \(\|cf\| = |c|\|f\|\). 3. \(\|f+g\| \leq \|f\|+\|g\|\) (Triangle inequality). Let's check these properties for the given norm: 1. By definition, both maximums are non-negative (as they come from taking the absolute value), so the norm is always non-negative. If \(f=0\), then \(\|f\|=0\). Conversely, if \(\|f\|=0\), it means that \(f(x)=0\) and \(f'(x)=0\) for all \(x\in [a,b]\), hence \(f=0\). 2. For any scalar \(c\), \(\|cf\| = \max_{x \in [a, b]}|cf(x)|+\max_{x \in [a, b]}|c f'(x)| = |c|\max_{x \in [a, b]}|f(x)|+|c|\max_{x \in [a, b]}|f'(x)| = |c|\|f\|\). Property 2 is satisfied. 3. To verify the Triangle Inequality, note that \begin{align*} \|f+g\|&=\max_{x\in [a, b]} |f(x)+g(x)|+\max_{x \in [a, b]}|f'(x)+g'(x)| \\ &\leq \max_{x\in [a, b]} (|f(x)|+|g(x)|)+\max_{x \in [a, b]}(|f'(x)|+|g'(x)|)\\ &\leq \max_{x \in [a, b]}|f(x)|+\max_{x \in [a, b]}|g(x)|+\max_{x \in [a, b]}|f'(x)|+\max_{x \in [a, b]}|g'(x)| \\ &= \|f\|+\|g\|. \end{align*} Property 3 is satisfied, so the given function \(\|f\|\) is indeed a norm.

To show that \(C^{1}([a, b])\) is complete, we need to find a Cauchy sequence of functions \((f_n)_{n=1}^{\infty}\) in \(C^{1}([a, b])\) and prove that it converges to some function \(f \in C^{1}([a, b])\) with respect to the given norm. Let \((f_n)_{n=1}^{\infty}\) be a Cauchy sequence in \(C^{1}([a, b])\). This means that for every \(\epsilon > 0\), there exists an integer \(N\) such that \(\|f_n - f_m\| < \epsilon\) for all \(n, m > N\). Due to the norm definition, it follows that the sequences \((f_n)\) and \((f_n')\) of functions and their derivatives are Cauchy sequences in \(C([a,b])\) with the standard supremum norm. As \(C([a, b])\) is a Banach space, these sequences converge uniformly to their respective limits: \(f_n \to f\) and \(f_n' \to g\). To prove that \(f \in C^{1}([a, b])\), we need to show that \(f\) is differentiable and its derivative coincides with \(g\). We will show this by proving that \((f_n')\) converges uniformly to \(f'\). Let \(x \in [a, b]\). Then, \(|(f_n'(x)-f_m'(x))| \leq \|f_n-f_m\| \to 0\) as \(n, m \to \infty\). Hence, \((f_n'(x))\) is a Cauchy sequence in \(\mathbb{R}\) and converges to some limit, say \(h(x)\). Therefore \(g(x) = h(x)\). Now, we have \(\|f_n - f\| \to 0\) and \(\|g - h\| \to 0\) as \(n \to \infty\), which implies that \(\|f_n - f\| + \|g - h\| \to 0\) as \(n \to \infty\). Hence, \((f_n)_{n=1}^{\infty}\) converges to \(f\) in the \(C^{1}([a, b])\) space. Since every Cauchy sequence in \(C^{1}([a, b])\) has a limit in \(C^{1}([a, b])\), \(C^{1}([a, b])\) is complete. Therefore, \(C^{1}([a, b])\) with the given norm is a Banach space.