# Showing a certain pre-Hilbert space is complete

The following is problem 1.3 in chapter 1 of Conway's A Course in Functional Analysis.

Show the space $\mathcal H$ in Example 1.8 is a Hilbert space.

For reference, I give the text of Example 1.8. The field $\mathbb F$ can be either $\mathbb R$ or $\mathbb C$.

Let $\mathcal H=$ the collection of all absolutely continuous functions $f:[0,1]\rightarrow \mathbb F$ such that $f(0)=0$ and $f'\in L^2(0,1)$. If $\langle f,g \rangle =\int_0^1 f'(t)\overline{g'(t)}\ dt$ for $f$ and $g$ in $\mathcal H$, then $\mathcal H$ is a Hilbert space (Exercise 3).

• I didn't see this question on the site, but please mark it as a duplicate if has appeared before. – Potato Aug 3 '13 at 4:42
• Note, all Hilbert spaces are complete, so your title is a mistake - you are trying to prove that a space is a Hilbert space, and have (presumably) proven all the properties other than completeness. – Thomas Andrews Aug 3 '13 at 4:45
• @ThomasAndrews Indeed. I've updated the title. – Potato Aug 3 '13 at 4:47
• What do you mean by "I have posted this question for reference purposes"? This site is not meant to be a repository for your homework attempts, it is meant to be a question and answer site. – Thomas Andrews Aug 3 '13 at 4:47
• @ThomasAndrews See meta.math.stackexchange.com/questions/10411/… – Potato Aug 3 '13 at 4:48

Suppose $\{f_n\}$ is a Cauchy sequence in $\mathcal H$. Then $\{f'_n\}$ is a Cauchy sequence in $L^2[0,1]$. This implies it has a limit $g$, because $L^2$ is complete. We will show that the natural guess, $h(x)=\int_0^x g(t)\ dt$, is the limit of $\{f_n\}$ in $\mathcal H$. First, note this function is in $\mathcal H$: its derivative is in $L^2[0,1]$, and $h(0)=0$. It is the limit of $\{f_n\}$ in $\mathcal H$ essentially by definition:
$$\| f_n - h\|_{\mathcal H} = \|f_n' - g\|_{L^2[0,1]},$$
and the latter quantity tends to $0$ by the construction of $g$.
• You could also simply say that $H$ is isometrically isomorphic to $L^2$ via $f\longmapsto f'$, whose inverse, as you noted, is given by $g\longmapsto \int_0^xg$. So $H$ is a Hilbert space, without going through Cauchy again. – Julien Aug 3 '13 at 17:23
• Hi!, I don't understand why if you suppose $\{f_n\}$ is a Cauchy sequence in $\mathcal H$, then $\{f'_n\}$ is a Cauchy sequence in $L^2[0,1]$. Can anyone help me, thanks! – MathUser Sep 24 '16 at 17:45
• Since: $$\| f_n -f_m\|_{\mathcal H} = \|f_n'- f_m'\|_{L^2[0,1]}$$ – Amin Jan 15 '18 at 14:20