Consider the map $$i: (C^1[0,1], ||·||_{W^{1,2}}) → (C^0[0,1], ||·||_∞)$$ which maps every function to itself, and with Sobolev norm defined as $$||u||_{W^{1,2}}=||u||_{L^2}+||u'||_{L^2}.$$ Is $i$ linear, continuous, compact?


Consider $u,v\in C^1[0,1]$ and $a,b\in \mathbb{R}$:



By fundamental theorem of calculus: $u(x)=u(0)+\int_0^x u'(t)dt$. Then:

$\begin{align*} |u(x)| &\le |u(0)|+\int_0^x |u'(t)|dt \\ &\le |u(0)|+\int_0^1 |u'(t)|dt \\ &\le C|u(0)|^2+C\int_0^1 |u'(t)|^2dt \\ &\le C\int_0^1 |u(t)|^2dt+C\int_0^1 |u'(t)|^2dt \\ &\le C(||u||_{L^2}+||u'||_{L^2}) \\ &=C||u||_{W^{1,2}} \end{align*}$

for $C$ large enough and by mean value theorem.

Since $i$ is linear and bounded, it is also continuous.


$i$ is defined on an infinite-dimensional space, so by Riesz theorem the closed unit ball $B$ is not compact. If the dual norm of $i$ would be $1$, then we could say that $i(B)\subseteq B$, and so also $i$ would not be compact. But in this exercise I cannot show this is the case.

Are the computations for linearity and continuity correct?

How to check the compactness?

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    $\begingroup$ Do you know the Ascoli theorem? $\endgroup$ – Mindlack Jan 21 '19 at 14:59
  • $\begingroup$ It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right? $\endgroup$ – sound wave Jan 21 '19 at 15:11
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    $\begingroup$ The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous. $\endgroup$ – Mindlack Jan 21 '19 at 15:18
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    $\begingroup$ The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm. $\endgroup$ – Mindlack Jan 21 '19 at 15:40
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    $\begingroup$ Could, as in : not forbidden by Riesz theorem. $\endgroup$ – Mindlack Jan 21 '19 at 15:59

For compactness it has to be shown that a bounded set is relatively compact. Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f \in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} \le M$.

If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.

And in fact : Let $a, b \in [0,1]$, and $f \in G $. Then we have $ |f(a)-f(b)| = |\int_a^b f' | = |\int_a^b (f' \times 1) | \le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} \le ||f'||_{L^2}|a-b|^{1/2} \le M|a-b|^{1/2}.$

So Arzela-Ascoli is applicable and we have proven the requested property.

For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_\infty \le C (||u||_{L^2}+||u'||_{L^2})$. Now let us pick some $u$. Then there is a $r\in [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $x\in [0,1]$ we have $|u(x)| = |\int^x_r u'(t)dt +u(r)| \le |\int^x_r u'(t)dt| +|u(r)| = |\int^x_r u'(t) \times 1 dt| +|\int_0^1 |u(r)| \times 1 dt| \le ||u'||_{L^2} + |\int_0^1 |u(t)| \times 1 dt|\le ||u'||_{L^2} + ||u||_{L^2}$

Thus $C=1$ and we are done.

  • $\begingroup$ Thank you very much! Could you also say if the computation about continuity is correct? $\endgroup$ – sound wave Jan 21 '19 at 17:13
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    $\begingroup$ Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$. $\endgroup$ – Maksim Jan 21 '19 at 17:17
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    $\begingroup$ See edit above. $\endgroup$ – Maksim Jan 21 '19 at 17:53
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    $\begingroup$ That is needed for the inequality $|\int_0^1 |u(r)| \times 1 dt| \le |\int_0^1 |u(t)| \times 1 dt|$. Note that the first integrand is constant but the second is not. $\endgroup$ – Maksim Jan 21 '19 at 18:05
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    $\begingroup$ $|u(r)| =|u(r)| \times (1-0) = |u(r)|\int_0^1 dt= \int_0^1 |u(r)| dt =|\int_0^1 |u(r)| dt|$ $\endgroup$ – Maksim Jan 28 '19 at 10:49

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