The space $C^1[a,b]$ with respect to the norm $\int ^a _b |f| +\int ^a _b |f'|$ is not complete

Question

I am trying to find a Cauchy sequence that does not converge in this space. My attempt is $f_n = \sqrt{x^2+\frac{1}{n}}$, but I do not know how to prove it. Also, what would be the completion of this space? I have the feeling that it should be the space of absolute continuous functions since we need something that is differentiable, but I still do not know how to prove this.

Answer 1

I'll assume you know $L^1[a,b]$ and $C[a,b]$ are complete spaces.

Let $\mathcal{AC}[a,b]$ denote the set of absolutely continuous functions on $[a,b]$, equipped with the norm $$ \|f\| = \int_{a}^{b}|f(t)|+|f'(t)|dt. $$ First I'll show you that $\mathcal{AC}[a,b]$ is complete. To this end, note that $$ (x-a)f(x) = (t-a)f(t)|_{t=a}^{x}=\int_{a}^{x}f(t)+(t-a)f'(t)dt \\ (x-b)f(x) = (t-b)f(t)|_{t=b}^{x}=\int_{b}^{x}f(t)+(t-b)f'(t)dt. $$ Subtracting the two gives $$ (b-a)|f(x)| \le \int_{a}^{b}|f(t)|+(b-a)|f'(t)|dt \le(1+(b-a))\|f\|\\ \|f\|_{C[a,b]} \le \frac{1+(b-a)}{b-a}\|f\|. $$ Therefore, if $\{ f_n \}\subset\mathcal{AC}[a,b]$ is a Cauchy sequence, then $\{ f_n \}$ is a Cauchy sequence in $C[a,b]$, and $\{ f_n'\}$ is a Cauchy sequence in $L^1[a,b]$. Hence, $f_n$ converges uniformly to some $f\in C[a,b]$ and $\{f_n'\}$ converges to some $g\in L^1$. Then, $$ f_n(x)=f_n(a)+\int_{a}^{x}f_n'(t)dt \implies f(x)=f(a)+\int_{a}^{x}g(t)dt, $$ which implies that $f$ is absolutely continuous with $f'=g$ a.e.. So, \begin{align} \|f_n-f\| & \le \|f_n-f\|_{L^1}+\|f_n'-g\|_{L^1} \\ & \le (b-a)\|f_n-f\|_{C[a,b]}+\|f_n'-g\|_{L^1}\rightarrow 0 \end{align} So $\mathcal{AC}[a,b]$ is complete.

To show that $\mathcal{AC}[a,b]$ is the completion of your space, you need some sort of density argument for $L^1[a,b]$. For example, suppose $f\in\mathcal{AC}[a,b]$, and suppose you know there is a sequence of polynomials $\{ p_n \}$ such that $\|f'-p_n\|_{L^1[a,b]}\rightarrow 0$ as $n\rightarrow\infty$. Then, define $$ g_n(x)=f(a)+\int_{a}^{x}p_n(t)dt. $$ Next we show that $\{ g_n \}\subset\mathcal{AC}[a,b]$ converges to the given $f\in \mathcal{AC}[a,b]$ in the norm of $\mathcal{AC}[a,b]$, which proves your space is dense in $\mathcal{AC}[a,b]$, and, thus, is the completion of your space $X$. To do this, note that $$ |f(x)-g_n(x)| = \left|\int_{a}^{x}(f'-p_n)dt\right| \le \|f'-p_n\|_{L^1} \\ \|f-g_n\|_{L^1} \le (b-a)\|f'-p_n\|_{L^1}. $$ Therefore, \begin{align} \|f-g_n\| & \le \|f-g_n\|_{L^1}+\|f'-g_n'\|_{L^1} \\ & \le \{(b-a)+1\}\|f'-p_n\|_{L^1}\rightarrow 0. \end{align} Therefore, $\mathcal{AC}[a,b]$ is complete; your space $X$ has the $\mathcal{AC}[a,b]$ norm; and $X$ is dense in $\mathcal{A}[a,b]$. So $\mathcal{AC}[a,b]$ is the completion of your space $X$.