# Is $i: (C^1, ||·||_{W^{1,2}}) → (C^0, ||·||_∞)$ a linear, continuous, compact map?

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?

Linearity

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

$$i(au+bv)=au+bv=ai(u)+bi(v)$$

Continuity

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.

Compactness

$$i$$ 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?

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

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