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Is $C^1[a,b]$ with the norm $\left \| f \right \|_1=(\int_{a}^{b}\left | f(t) \right |dt)+(\int_{a}^{b}\left | f´(t) \right |dt)$ a complete space?

I thought with parabolas based on this link , but the area is infinite $C([0, 1])$ is not complete with respect to the norm $\lVert f\rVert _1 = \int_0^1 \lvert f (x) \rvert \,dx$.

Thanks.

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mathreda
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    +1 for having searched the site, and for contributing to its coherence by linking! – Bart Michels Sep 18 '16 at 20:43
  • The derivative is $L^1$ but it doesn't have to be continuous. So find a sequence of continuous functions converging to $1_{x < a+1}$ in the $L^1$ norm, or to $(x-a)^{-1/2}$ – reuns Sep 19 '16 at 04:13
  • As an aside, you have asked a number of questions on this site and received many responses but you have accepted none of the answers. The general protocol with these sites is that you accept the best (in your opinion) answer (assuming it answers your question satisfactorily). – copper.hat Sep 19 '16 at 23:07

2 Answers2

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To avoid abusing notation, let $\|\cdot \|_1$ represent the usual $L^1$ norm and $\|f\|_* = \|f\|_1 + \|f'\|_1$.

Consider $C^1[-1,1]$ with the above norm, and let $g_n(x) = {2 \over \pi}\arctan nx$. Note that the $g_n$ are smooth, odd, strictly monotonically increasing, $g_n(0) = 0$, $|g_n(x) \le 1$ for any $x$ and if $x \neq 0$, $\lim_n g_n(x) = \mathbb{sgn} \ x$. The discontinuity of the limit function at $x=0$ is the crucial element here.

If $\epsilon>0$, $m \le n$ and $x \ge \epsilon $, then $|g_n(x)-g_m(x)| \le |1-g_m(x)| \le |1-g_m(x)| \le |1-g_m(\epsilon)|$. Then $\|g_n-g_m\|_1 = 2\int_0^1 |g_n(x)-g_m(x)| dx \le 2 \epsilon + 2 |1-g_m(\epsilon)|$, and so $g_n$ are Cauchy in the $\| \cdot \|_1$ norm.

Define $f_n(x) = \int_0^x g_n(t)dt$, and note that $f_n' = g_n$. Note that the $f_n$ are smooth, even and hence $f_n'(0) = 0$. It is not hard to show, but irrelevant to this proof, that $\lim_n f_n(x) = |x|$.

Note that $|f_n(x)-f_m(x) | \le \int_{-1}^1 |g_n(x) - g_m(x)| dx = \|g_n-g_m\|_1$, hence $\|f_n-f_m\|_1 \le 2 \|g_n-g_m\|_1$ and so, for any $ \epsilon>0$, we have $\|f_n-f_m\|_* \le 3 (2 \epsilon + 2 |1-g_m(\epsilon)|) $ from which it follows that $f_n$ are Cauchy.

Then $f_n$ is Cauchy, but has no limit in $C^1[-1,1]$. Very roughly, it has no limit with respect to $\|\cdot \|_*$ because the only continuous function to which the $f_n$ can converge is $x \mapsto |x|$ and this fails to be differentiable, let alone continuously differentiable, at $x=0$.

To show that there is no limit in $C^1[-1,1]$, we suppose $\|f_n -f\|_* \to 0$ with $f \in C^1[-1,1]$ look for a contradiction.

First, note that $f$ must be even, since the $ f_n$ are. To see this, let $\phi(x) = f(-x)$ and note that $\|f-\phi\|_1 \le \|f-f_n\|_1 + \|f_n-\phi\|_1 = 2 \|f-f_n\|_1$, and since $n$ is arbitrary, we have $\|f-\phi\|_1 = 0$. Since $f, \phi$ are continuous, we have $f= \phi$ and so $f$ is even.

Since $f$ is even, we have $f'(0) = 0$. There is some $\delta>0$ such that if $|x| < \delta$, then $|f'(x)| < {1 \over 2}$. However, $f_n'({\delta \over 2})=g_n({\delta \over 2}) \to 1$, which is a contradiction.

copper.hat
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For the sake of better notation, let's use $\Vert\cdot\Vert_1$ instead to denote the norm $$\Vert f\Vert_1=\int_a^b|f(t)|dt$$ on $C^0[a,b]$, and $\Vert\cdot\Vert_{1,1}$ instead to denote the norm you are considering: $$\Vert f\Vert_{1,1}=\Vert f\Vert_1+\Vert f'\Vert_1.$$

Take $f_n$ as in the link you provided, and set $F_n(t)=\int_a^t f_n(s)ds$. Note that \begin{align*} \Vert F_n-F_m\Vert_1\leq\int_a^b\int_a^t|f_n(s)-f_m(s)|dsdt=\int_a^b\int_s^b|f_n(s)-f_m(s)|dtds \end{align*} by Fubini, and then $$\Vert F_n-F_m\Vert_1\leq\int_a^b(b-s)|f_n(s)-f_m(s)|ds\leq(b-a)\Vert f_n-f_m\Vert_1$$ so $\Vert F_n-F_m\Vert_{1,1}\leq (1+b-a)\Vert f_n-f_m\Vert_1$, so $(F_n)$ is Cauchy in $(C^1[a,b],\Vert\cdot\Vert_{1,1})$.

But the derivative map $D:(C^1[a,b],\Vert\cdot\Vert_{1,1})\to(C^0[a,b],\Vert\cdot\Vert_1)$ is continuous, and $D(F_n)=f_n$ does not converge, so $F_n$ does not converge as well.

EDIT: Let's explain the application of Fubini with some more details. The usual version of Fubini is that $\int_a^b\int_c^df(s,t)dsdt=\int_c^d\int_a^bf(s,t)dtds$.

Now suppose we have an integral of the form $\int_a^b\int_a^tf(s,t)dsdt$, where $f:[a,b]\times[a,b]\to\mathbb{R}$.

Let $g(s,t)=\begin{cases}1&\text{ if }s\leq t\\0&\text{ otherwise}\end{cases}$.

When we fix $t$, the inner variable of the inner integral ranges from $a$ to $t$, and so it can be rewritten as $$\int_a^tf(s,t)ds=\int_a^b g(s,t)f(s,t)ds$$ So we can rewritte the double integral as $$\int_a^b\int_a^tf(s,t)dsdt=\int_a^b\int_a^bg(s,t)f(s,t)dsdt$$ Now apply Fubini: $$\int_a^b\int_a^tf(s,t)dsdt=\int_a^b\int_a^bg(s,t)f(s,t)dtds$$ Loog at the inner integral in the RHS, i.e., for a fixed $s$, we have $$\int_a^b g(s,t)f(s,t)dt=\int_s^bf(s,t)$$ by definition of $g$, and therefore $$\int_a^b\int_a^tf(s,t)dsdt=\int_a^b\int_s^bf(s,t)dtds$$ We applied the above to $f(s,t)=|f_n(s)-f_m(s)|$.

Luiz Cordeiro
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  • I dont understand the inequalities, by fubini $\int_{a}^{b}\int_{a}^{t}dsdt=\int_{a}^{t}\int_{a}^{b}dtds$, or ? – mathreda Sep 19 '16 at 02:13
  • @mathreda Note that $\int_a^t(\int_a^bdt)ds$ does not make sense, because $t$ is a variable of integration in the inner integral, and so cannot be a limit of integration in the outer integral. I've added some details. – Luiz Cordeiro Sep 19 '16 at 04:11
  • I don't even understand what you are trying to show. Find instead a Cauchy sequence whose limit is not $C^1$. – reuns Sep 19 '16 at 06:35
  • blah blah. Show a sequence of $C^1([0,1])$ functions that is Cauchy for the $|f|{L^1([0,1])}+|f'|{L^1([0,1])}$ norm and whose limit isn't $C^1$ – reuns Sep 19 '16 at 07:09
  • Cordeiro, integrate not implies continuity, for example, in $F_1=0$ if x lives in [0,1/2], $F_1=(x^2-x)/2$ if x lives in [1/2, a_m] and $F_1=x$ if x lives in [a_m,1]. – mathreda Sep 20 '16 at 23:34
  • @mathreda No. It is standard from calculus that a integral $F(x)=\int_a^x f(t)dt$ of a continuous function $f$ is differentiable (in particular continuous), and the derivative is $F'=f$. You got confused in the integration, because $f_n$ is defined by parts. Instead (considering $[a,b]=[0,1]$ as in the link) we have $F_n(x)=0$ for $x\leq 1/2$, $F_n(x)=\int_{1/2}^x n(t-1/2)dt=\frac{nx^2}{2}-\frac{nx}{2}-\left(\frac{n(1/2)^2}{2}-\frac{n(1/2)}{2}\right)$ for $1/2\leq x\leq 1/2+1/n$, and of the form $F_n(x)=x+c$ for some constant $c$ (which you can calculate) for $1/2+1/n\leq x$. – Luiz Cordeiro Sep 21 '16 at 01:48
  • I see, thanks. Please how is k such that $n,m\geq k$ is Cauchy in $\left | F_{n}-F_{m} \right |{1,1\leq }(1+b-a)\left | f{n}-f_{m} \right |_{1} $? – mathreda Sep 21 '16 at 06:30
  • Hi Cordeiro, please how can I show this: $\int_{0}^{x}\int_{0}^{x}f(y)dydy=\int_{0}^{x}(x-y)f(y)dy$ – mathreda Apr 05 '17 at 05:24