# $||f||_1 =(\int_a^b [|f|^2+|f'|^2]dx)^{1/2}$. Is this normed space complete?

Define $C_1^1[a,b]$ to be the space of continuously differentiable functions on $[a,b]$, with norm

$$||f||_1 =\left(\int_a^b \left(|f|^2+|f'|^2\right) dx \right)^{1/2}$$

Is this normed space complete?

So far I have shown that this is a normed space by satisfying the four axioms of a normed space. Now I need to show that it is complete. That is, show that every Cauchy sequence in this space is convergent. I'm stuck on how to do this part. Any hints or solutions are greatly appreciated.

• Try showing that it is not complete. Since the space is not complete under that norm, that has higher chances of success. – Daniel Fischer Jun 16 '15 at 21:08
• hmmm ok, so a counter-example – 1233dfv Jun 16 '15 at 21:21
• And it would be sporting to note that this is the $H^1$ ($L^2$) Sobolev norm, so the completion is denoted/called the Sobolev space $H^1[a,b]$. Not to recommend googling rather than finding your own counter-example, but for follow-up later. – paul garrett Jun 16 '15 at 21:23
• I can't figure out a counter-example. Any recommendations or links? – 1233dfv Jun 16 '15 at 21:57
• Hint: concentrate on the derivative. You want that to have a discontinuous $L^2$-limit. – Daniel Fischer Jun 16 '15 at 23:40

Hint

Pointwise limit of differentiable functions needn't be differentiable.

• I think that my answer is a hint to find a counterexample, and this is precisely my intention. A pointwisely convergent sequence of functions of $C_1^1[a,b]$ will be a Cauchy sequence with this norm, but if the limit function is not differentable, the sequence will not converge in $C_1^1[a,b]$. – ajotatxe Jun 16 '15 at 23:59
• Pointwise convergence does not imply $L^2$-convergence, much less $L^2$-convergence of the derivatives. – Daniel Fischer Jun 17 '15 at 8:57

I was working on a similar problem. Here is a way of showing that $C^1[a,b]$ is not a Banach space. We will assume $a=0$ and $b=1$; the method behind the proof below will motivate the general idea. Let $\{f_j\}_{j\geqslant 1}\subset C^1[0,1]$ be a sequence, defined as follows.

$$f_j(t)=\left\{ \begin{array}{ll} \frac{1}{2(j+1)}(2t)^{(j+1)}, ~~0\leqslant t<\frac{1}{2}\\ (t-\frac{1}{2})+\frac{1}{2(j+1)}, ~~\frac{1}{2}\leqslant t\leqslant 1 \end{array} \right.$$

It's easy to check that $f_j$ is differentiable on $(0,1)$ and continuous on $[0,1]$. Now we claim that $\{f_j\}_{j\geqslant 1}$ is Cauchy.

Proof. Check that for any $m,n\in\mathbb{N}$ (assuming $m>n$),

\begin{eqnarray} d(f_m,f_n) & = & \Vert f_m-f_n\Vert \\ & = & \left(\int_0^1(f_m-f_n)^2+[(f_m-f_n)']^2dt\right)^{\frac{1}{2}} \\ & = & \left(\int_0^{1/2}(f_m-f_n)^2+[(f_m-f_n)']^2dt+\int_{1/2}^1(f_m-f_n)^2+[(f_m-f_n)']^2dt\right)^{\frac{1}{2}}. \end{eqnarray}

The two integrals in the squareroot are easily bounded above by $\left[\frac{1}{2(n+1)}\right]^2+\left(\frac{1}{2}\right)^{2n+1}$, which approaches zero. Hence, for any $\varepsilon>0$ picked, there exists $N\in\mathbb{N}$ such that for all $m,n\geqslant N$ the expression above is strictly bounded above by $\varepsilon$. $\blacksquare$

We claim that $\{f_j\}_{j\geqslant 1}$ does not converge to a member of $C^1[0,1]$. What is $f=\lim_{j\rightarrow\infty}f_j$? It is given by

$$f(t)=\left\{ \begin{array}{ll} \lim_{j\to\infty}\frac{1}{2(j+1)}(2t)^{(j+1)}, ~~0\leqslant t<\frac{1}{2}\\ \lim_{j\to\infty}(t-\frac{1}{2})+\frac{1}{2(j+1)}, ~~\frac{1}{2}\leqslant t\leqslant 1 \end{array} \right.$$

In other words, it's the function $f(t)=0$ when $0\leqslant t<\frac{1}{2}$ and $f(t)=t-\frac{1}{2}$ when $\frac{1}{2}\leqslant t\leqslant 1$. It's straightforward to check that $f$ isn't differentiable on $(0,1)$.

• Your argument is true, but you use the pointwise limit to calculate the limit function $f$. Is that correct? Shouldn't it be the convergence in norm, that is, a function $f$ will be the limit of the sequence considered when $\Vert f_j-f\Vert\rightarrow 0$ and not when $f=\lim f_j$. – Senna Mar 7 at 12:24