Showing that space of absolutely continuous functions is Banach space

Let $$AC([a, b])$$ denotes space of absolutely continuous functions such that $$f:[a, b] \mapsto \mathbb{R}$$ for all $$f \in AC([a, b])$$.

Let's define a norm on this space in the following way $$\lvert\lvert f \rvert \rvert = \int \limits_{a}^{b} \lvert f(x) \rvert + \lvert f'(x) \rvert \, dx.$$

I would like to show that $$AC([a, b], \lvert\lvert \cdot \rvert \rvert)$$ is a Banach space.

Searching for a candidate for the limit of a Cauchy's sequence
Let $$f_n$$ be a Cauchy sequence in $$AC([a, b])$$ and fix $$\epsilon > 0$$. Thus for each $$x \in [a, b]$$ we do have $$\lvert f_n(x) - f_m(x) \rvert < \epsilon$$ and $$\lvert f_n'(x) - f_m'(x) \rvert < \epsilon$$ for $$m, n > N_0$$.
Because $$\mathbb{R}$$ is complete thus $$f_n(x)$$ is pointwise convergent to $$f(x)$$ and $$f_n'(x)$$ is pointwise convergent to $$f'(x)$$. So we found our candidate.

Convergence in norm
Let $$N_1$$ be such that $$\lvert f(x) - f_n(x) \rvert < \frac{\epsilon}{2(b-a)}$$ and for $$N_2$$ we have $$\lvert f'(x) - f_n'(x) \rvert < \frac{\epsilon}{2(b-a)}$$. Let $$N_0 = \max\{N_1, N_2 \}$$ and it's obvious that $$\lvert\lvert f - f_n \rvert \rvert < \epsilon$$.

Now we have to show that $$f \in AC([a, b])$$
$$\sum \limits_{k = 1}^{N} \lvert f(x_k) - f(y_k) \lvert \le \sum \limits_{k = 1}^{N} \big( \lvert f(x_k) - f_n(x_k) \lvert + \lvert f_n(x_k) - f_n(y_k) \lvert + \lvert f(y_k) - f_n(y_k) \lvert \big) < \epsilon$$ Now we can go with $$N \to \infty$$ and we do have what we wanted. The same for derivatives.

Is my proof true? I'm not sure whether everything I did with derivatives is done correctly.

I would appreciate any comments, hints or tips.

• The usual norm on the "space of absolutely continuous function", better known as $\operatorname{L}_1$, is $\lVert f \rVert_1 := \int |f(x)| dx$. – Olivier Roche Dec 12 '19 at 14:23
• @OlivierRoche Unfortunately I was forced to use to norm in my post. – Hendrra Dec 12 '19 at 15:17
• But what if $f$ isn't differentiable?What's $f'$ then? – Olivier Roche Dec 12 '19 at 17:12
• @OlivierRoche Absolutely continuous functions are differentiable almost everywhere which is sufficient to define the norm used in this question – Rhys Steele Dec 12 '19 at 19:18
• @Hendrra How are you concluding that there is some $N_0$ such that for $x \in [a,b]$, $|f_n(x) - f_m(x)| < \varepsilon$ for $n,m \geq N_0$? This is a pointwise statement but your norm involves integrals of $f$ and $f'$. – Rhys Steele Dec 12 '19 at 19:22

First note that if a sequence $$\{f_n\}_{n\in\mathbb{N}}\subset AC([a,b])$$ is a cauchy sequence in the sensee of your norm, then $$\{f_n\}_{m\in\mathbb{N}}$$ and $$\{f_n^{'}\}_{m\in\mathbb{N}}$$ are cauchy sequences in the sense of the $$L^1$$ norm, so you can argue the existence of functions $$f$$ and $$g$$ such that $$f_n\rightarrow f$$ and $$f_n^{'}\rightarrow g$$ in the sense of the $$L^1$$ norm. Now observe that \begin{align} f(x)=&\lim_{n\rightarrow\infty}f_{n_k}(x)\\ =&\lim_{n\rightarrow\infty}\int_a^x f_{n_k}'(t)dt\\ =&\int_a^x\lim_{n\rightarrow\infty}f_{n_k}'(t)dt\\ =&\int_a^x g(t)dt \end{align} where you can argue the first equality almost everywhere noting that the convergence in $$L^p$$ implies the sonvergence $$a.e.$$ of some sub sequence, and the last two equalities using the theorem of dominated convergence. Finally, since the class of absolutely continuous functions is exactly the set of functions that can be represented as the integral of other function and $$f'=g$$ almost everywhere, this concludes the proof.

• Your first equality relies on having $f_n \to f$ almost everywhere but you only know that $f_n \to f$ in $L^1$. You need to pass to a subsequence that converges almost everywhere to fix this (minor) issue – Rhys Steele Dec 12 '19 at 21:45
• Thanks for your comment Rhys, I've just edited it and added a brief explanation about the need of the sub sequence. Thanks again. – Esteban Gutiérrez Dec 13 '19 at 0:21

You seem to be confusing your norm with pointwise convergence when finding your candidate.

Note that a function $$g\in AC([a,b])$$ if and only if there exists $$g'\in L^1$$ and some $$C$$ such that $$g(x)=\int_a^x g'(t)\textrm{d}t+ C$$ a.e.. Hence, picking a Cauchy sequence $$(f_n)_{n\in\mathbb{N}}\subseteq AC([a,b]),$$ we let $$C_n$$ be constants as before.

Note that for $$n,m\in \mathbb{N}$$ we have $$|C_n-C_m|(b-a)=\left|\int_a^b f_n(x)-f_m(x)\textrm{d}x-\int_a^b f'_n(x)-f'_m(x)\textrm{d}x\right|\leq \|f_n-f_m\|,$$ implying that the $$(C_n)_{n\in \mathbb{N}}$$ form a Cauchy sequence in $$\mathbb{R}$$, hence has a limit $$C_{\infty}$$.

Now, $$\| f_n'-f_m'\|_{L^1}\leq \| f_n-f_m\|$$ as well, so by completeness of $$L^1([a,b]),$$ $$f_n'$$ has some limit $$g_{\infty}$$.

Furthermore, denote $$h(x)=\int_a^xg_{\infty}(t)\textrm{d}t+C_{\infty}$$. Then,

$$\| f_n-h(x)\|\leq |C_{\infty}-C_n|+\int_a^b\int_a^x |f_n'-g_{\infty}|\textrm{d}x\textrm{d}t\leq |C_{\infty}-C_n|+(b-a)\| f_n'-g_{\infty}\|_{L^1},$$ which goes to $$0$$. Hence, $$f_n\xrightarrow{\|\cdot\|} h$$ and we get that $$AC([a,b])$$ is complete.

Note that the only reference to pointwise convergence that we actually need lies way back in the proof of completeness of the $$L^p$$ spaces. Note also that the candidate limit that we consider is already a priori absolutely continuous, so this causes us no issue.