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Show that the set of the functions $A_M:=\{f ∈ C^{k+1}([0, 1]) : \|f\|_{C^{K+1}} ≤ M\}$ is compact in $C^{k}[0,1]\ \ \forall M \geq 0$.

N.B.:
$$\| f\|_{C^{K+1}}=\|f\|_{C^{0}}+\|f^{(k+1)}\|_{C^{0}}$$

I started by showing that $C^{k}[0,1]$ is complete with the $C^{k}$ norm.
I'd like to use this to show that the set is complete.
Then prove that it is totally bounded and so compact.
But I don't have an idea of how to do it...

Thanks.

Lance
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  • I think you have a mistake on $M \leq 0$. – Needmoremath Feb 02 '20 at 17:39
  • you are right, is >= 0 – Lance Feb 02 '20 at 19:10
  • If the derivative of a sequence is bounded then the Lipschitz constant is bounded and the sequence is equicontinuous. Now use Azerla-Ascoli to get compactness. – s.harp Feb 02 '20 at 19:55
  • Are you sure about the indices? Why $k+1$-st norm uses $k$'th derivative?! What would $C_1$norm be? $2 \times C_0 $ norm of $f$? – Behnam Esmayli Feb 08 '20 at 03:28
  • edited but the problem still remains – Lance Feb 08 '20 at 11:02
  • Take a bounded sequence ${f_n}_{n \in \mathbb{N}}$ in $A_M$. Then, $$ \forall n \quad |f_n|_0 \leq M, \quad |f_n^{(k+1)}|_0 \leq M , . $$

    Consider the sequence of functions ${f_n^{k}}_{n\in \mathbb{N}}$. Then, by mean value theorem, they are Lipschitz with Lipschitz constant uniformly bounded by $M$, more precisely, $$ \forall n \ \ \forall x \ \ \forall y: \quad |f^k_n(x) - f^k_n(y)| \leq M |x-y| , . $$ So, ${f_n^k}_n$ is equicontinuous. I am yet unable to prove equibounedness, if I do, then by ascolli-arzella a subsequence of $f^k_n$ would converge. Now repeat.

     – Behnam Esmayli Feb 12 '20 at 00:33
  • Another idea, which can give an elegant and immediate solution, is to show that the norm above is equivalent to the norm $$ |f|_0 + |f'|_0+|f''|_0+\cdots+|f^{(k+1)}|_0 , $$ a fact frequently used in Sobolev spaces. Then it immediately follows that if $f_n$ is bounded in this norm then $f^j_n$ is equibounded and equicontinuous for all $j\leq k$. find a convergent subseq for $j=k$, out of this one afurther subsequence for $j=k-1$, etc. The resulting subseq converges in all $|\cdot ^j|_0$$ norms for $0\leq j \leq k$$. – Behnam Esmayli Feb 12 '20 at 00:44
  • I tried by showing that the norm is equivalent to the classic ck norm but it works only with k = 2... I'm starting to think that the prof has given us the wrong definition of ck norm (this is a fall course of the first year! The solution should be easier ..) – Lance Feb 12 '20 at 09:49

2 Answers2

1

Start with this statement:

If $f_n :[0,1]\to\Bbb R$ bounded wrt $\|\cdot\|_\infty$ with $f_{n}^{(1)}$ being unbounded, then all derivatives (if they exist) $f_n^{(k)}$ are unbounded in $n$ wrt $\|\cdot\|_\infty$.

Let $\|f_n\|_\infty < C$. I will carry out the first few steps so that the induction argument becomes obvious.

If $f_n^{(1)}$ is unbounded pass to a subsequence to assume $\|f_n^{(1)}\|_\infty > n$. This means that $|f_n^{(1)}|$ must take on a value $>n$, however it is continuous and the integral of $f$ cannot be larger than $C$ as such when it reaches $n$ it has only a time $\frac{C}{n/2}$ to get back under $n/2$ ($n$ large enough for this number to be less than $1/2$). This means that the Lipschitz constant of $f^{(1)}$ is larger than $\frac{n^2}{2C}$. As such the derivative $f^{(2)}_n$ is unbounded as well, satisfying $\|f_n^{(2)}\|_\infty > \frac{n^2}{2C}$. Further $|f^{(2)}_n|$ cannot be larger than $\frac{n^2}{4C}$ on intervals of length $n\frac{8C^2}{n^2}+\frac{2C}n=\frac{8C^2+2C}n$ as otherwise $|f_n^{(1)}|$ would be larger than $n$ on an interval of length $\frac{2C}n$, which is not allowed.

You must now continue in this way. You must find that $f^{(k)}$ is both unbounded but cannot be "big" on an interval of length $> F^k(n)$ where $F^k(n)$ becomes small. This implies that the next derivative is unbounded in $n$, as the Lipschitz constant must be big, and that the next derivative cannot be big on any interval of length $F^{k+1}(n)$ which again becomes small, as otherwise $f^{(k)}$ cannot leave its "big derivative regime" quickly.

The above statement puts you in the position to use Azerla-Ascoli. If $\|f_n\|_\infty < C$ and $\|f_n^{(k+1)}\|_\infty < C$ you must find that all derivatves of order $≤k$ must be bounded (otherwise contradicting the above statment). Azerla-Ascoli tells you that any bounded equicontinuous sequence in $C([0,1])$ will admit a convergent subsequence.

If $f_n^{(k)}$ is bounded and its derivative $f_n^{(k+1)}$ is bounded then the $f_n^{(k)}$ are equicontinuous because they share a common Lipschitz constant. Azerla-Ascoli gives you the convergent subsequence for the $k$-th derivative. Continue in this way to find a sub-sequence so that all derivatives $≤k$ converge. Then make use of the following lemma:

If $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to $g$ then $f$ is differentiable with $f'=g$.

To find that what you got actually converges in the topology you are looking at.

s.harp
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  • didn't you missed to add the constant of integration? With your method we can prove that each f(n) is bounded in C^(k) but we can't conclude that there is a bound which works for every function... – Lance Feb 06 '20 at 11:09
  • You are right, what one has is $f_n^{(1)}(x) = \int_0^x f_n^{(2)}(t)dt + f_n^{(1)}(0)$. The first term is bounded by the obvious caclulation in the answer. The second term must also be bounded in $n$, for suppose it were unbounded and by passing to a subsequence suppose $f_n^{(1)}(0)≥ n$. Then you get an $\epsilon$ so that $f_n^{(1)}(x)≥\frac{n}2$ for all $x\in[0,\epsilon]$ and all $n$ (by using that $f^{(2)}$ is bounded), hence $|f_n(0)-f_n(\frac\epsilon2)|≥ \epsilon\frac n4$, contradicting boundedness of $f_n$. – s.harp Feb 06 '20 at 19:11
  • you're using the fact, if i'm right than $\epsilon$ doens't depend on $n$ because the second derivative is bounded... but you can do this trick only in $C^{2}$. With higher $k$ you can't say epsilon does't depend on n – Lance Feb 07 '20 at 15:44
  • I think basic idea in this answer is correct but we need to work backwards! Assuming $k$'th derivatives are bounded why are $k-1$-st derivatives an equibounded and equicontinuous collection of continuous functions? If this is verified then by A_A a convergentsubsequence will be found. – Behnam Esmayli Feb 08 '20 at 03:30
  • I tried using induction backward but it doesn't work... – Lance Feb 08 '20 at 12:41
  • I rewrote the answer, it now sketches how the induction will work to get that all derivatives are bounded. – s.harp Feb 08 '20 at 17:03
0

The set $A_M$ is not compact in $C^k[0,1]$. It is pre-compact, i.e., it is totally bounded, but not complete.

For example, for $k=0$, and $M=3$, then $$ f_n(x)=\sqrt{(x-1/2)^2+1/n^2}\in A_M\subset C^1[0,1], $$ and $f_n\to |x-1/2|=f$, in the $C[0,1]$ norm. But $f\not\in A_M$.

Yiorgos S. Smyrlis
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  • Nice example! So instead of looking at $C^{k+1}$ one should look at something like $L^{k+1}$, the space of $k$-times differentiable functions with the $k$-th derivative being Lipschitz continuous and take as norm $|f|_\infty + L(|f^{k}|)$ with $L(|f^{k}|)$ being the smallest Lipschitz constant of $|f^k|$. The Lipschitz-property should be preserved by any limit in $C^k$ so this set should be closed in $C^k$. – s.harp Feb 12 '20 at 19:10