# Prove $C_b^j (I, \mathbb{R})$ is a complete space

Let I be a real interval (possibly infinite).

Let $C_b^j (I, \mathbb{R}) =${$f:I \rightarrow \mathbb{R}$ such that $f$ is j-times continuously differentiable and $f^{(m)}$ is bounded for $m \leq j$}

Define a norm on this space to be

$$||f||_{C^j} = \sum_{m=0}^j || f^{(m)}||_{sup}$$

I want to prove that this space (with this norm) is complete. I know that any metric space can be naturally completed, so WLOG, if {$f_n$} is any Cauchy sequence in $C_b^j (I, \mathbb{R})$, we know there exists a limit function $f = lim_{n \rightarrow \infty} f_n$. The task is to show that f is a member of $C_b^j (I, \mathbb{R})$.

However, here I am having technical issues. If I want to show that f is bounded, continuous, or differentiable (j times), then I end up needing epsilon delta arguments based on the distance between $f$ and some $f_n$. However, the distance between $f$ and $f_n$ is based on the norm which already assumes that f is j times differentiable and that the derivatives are bounded. Can anyone shed some light on this circular bit of logic?

The case $j=0$ should be clear. For $j=1$ note that $J:C_b^1(I,\mathbb R) \to C_b(I,\mathbb R)^{2}$, $f\mapsto (f,f')$ is an isometry onto its range (if the codomain is equipped with the sum of the uniform norms of the components). Since $X^2$ is Banach for a Banach space $X$ and closed subspaces are complete, it is enough to show that the range of $J$ is closed. To see this check that $(f,g)$ is in the range if and only $f(x)-\int_{a}^xg(t)dt$ is constant (where $a\in I$ is fixed). Written differently, Range$(T)=$Kern$(S)$ for $S(f,g)(x)=f(x)-f(a)-\int_a^xg(t)dt$. The continuity of $S: C_b^1(I,\mathbb R)^2\to C_b^1(I,\mathbb R)$ then implies that Kern$(S)$ is closed.
For $j\ge 2$ use induction.
• Still the case $j=1$: If $(f_n)_n$ is Cauchy in $C_b^1$ then $(f_n)_n$ and $(f_n')_n$ are both Cauchy in $C_b$ and hence convergent to $f$ and $g$, respectively. As $f_n(x)=f_n(a)+\int_a^xf_n'(t)dt$ you get from the uniform convergence $f_n'\to g$ that $f(x)=f(a)+\int_a^x g(t)dt$ which shows $f\in C_b^1$ with $f'=g$ so that $f_n\to f$ in $C_b^1$. Aug 31, 2017 at 14:08
• I'm not quite clear on your notation. What is $(f_n)_n$? Aug 31, 2017 at 15:49
• Shorthand for a sequence $(f_n)_{n \in \mathbb N}$ of functions. Aug 31, 2017 at 16:52