The completeness of function spaces often depends on controlling the right features. For $C^1[0,1]$, controlling only the function values is not enough; the derivatives must also be controlled uniformly. The metric in this lesson combines the uniform distance of functions and the uniform distance of derivatives. :::definition[Space of Continuously Differentiable Functions] Let $C^1[0,1]$ be the set of all real-valued functions $x:[0,1]\to\mathbb{R}$ such that $x$ is continuous on $[0,1]$ and $x'$ exists and is continuous on $[0,1]$. ::: :::theorem[Completeness of Continuously Differentiable Functions] Let $C^1[0,1]$ be equipped with $$\begin{equation*} d(x,y)=\sup_{t\in[0,1]}|x(t)-y(t)|+\sup_{t\in[0,1]}|x'(t)-y'(t)|. \end{equation*}$$ Then $C^1[0,1]$ is complete. ::: :::proof Given that $C^1[0,1]$ has the stated metric. To prove that it is complete. Let $\{x_n\}$ be Cauchy. Then $\{x_n\}$ and $\{x_n'\}$ are Cauchy in the uniform metric on $C[0,1]$. Hence $x_n\to x$ uniformly and $x_n'\to y$ uniformly for continuous functions $x$ and $y$. Since $$\begin{equation*} x_n(t)-x_n(0)=\int_0^t x_n'(s)\,ds, \end{equation*}$$ passing to the limit gives $$\begin{equation*} x(t)-x(0)=\int_0^t y(s)\,ds. \end{equation*}$$ Thus $x'=y$, so $x\in C^1[0,1]$. Also $d(x_n,x)\to 0$. Hence $C^1[0,1]$ is complete. $\square$ ::: :::solved-problem Show that $x_n=(1,1/2,\ldots,1/n,0,0,\ldots)$ converges in $\ell_2$ to $x=(1,1/2,1/3,\ldots)$. ::: :::solution Let $x_n$ and $x$ be as stated. For $k\leq n$, the coordinates agree. For $k>n$, the difference is $1/k$. Hence $$\begin{equation*} d(x_n,x)=\left(\sum_{k=n+1}^{\infty}\frac{1}{k^2}\right)^{1/2}. \end{equation*}$$ The tail of the convergent series $\sum 1/k^2$ tends to zero. Therefore $d(x_n,x)\to 0$. Hence $x_n\to x$ in $\ell_2$. $\square$ ::: :::theorem[Completeness of Bounded Continuous Functions] Let $(X,d)$ be a metric space and let $C_b(X,\mathbb{R})$ be the set of bounded continuous real-valued functions on $X$, equipped with the uniform metric. Then $C_b(X,\mathbb{R})$ is complete. ::: :::proof Given that $C_b(X,\mathbb{R})$ has the uniform metric. To prove that it is complete. Let $\{f_n\}$ be Cauchy. In the complete space of bounded functions, it converges uniformly to a bounded function $f$. To show continuity, fix $x_0\in X$ and $\epsilon>0$. Choose $N$ such that $|f-f_N|<\epsilon/3$ everywhere. Since $f_N$ is continuous at $x_0$, choose $\delta>0$ such that $d(x,x_0)<\delta$ implies $|f_N(x)-f_N(x_0)|<\epsilon/3$. Then $|f(x)-f(x_0)|<\epsilon$. Hence $f$ is continuous. Therefore $C_b(X,\mathbb{R})$ is complete. $\square$ ::: :::remark Completeness ensures that Cauchy sequences have limits inside the space being studied. For function spaces, the selected metric determines which features survive in the limit. ::: :::exercise [1] State the metric on $C^1[0,1]$. [2] Why does the proof identify the limit of the derivatives? [3] Why do the truncations of $(1,1/2,1/3,\ldots)$ converge in $\ell_2$? ::: :::answer [1] $d(x,y)=\sup|x-y|+\sup|x'-y'|$. [2] The fundamental theorem of calculus connects $x_n$ and $x_n'$ and passes to the uniform limit. [3] The squared distance is the tail of $\sum 1/k^2$. ::: :::call-to-action[Continue to Completion] subtitle: Next, learn how every metric space embeds into a complete metric space. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/completion-and-isometric-embedding button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/coordinatewise-and-uniform-convergence :::