Space with continuous bounded derivatives of all functions is complete

Question

I'm doing the following exercise:

Let $I=(a,b)\subset \mathbb{R}$, and we call $H^k(I)$ the space of continuous functions on $I$ such that its derivatives are all bounded $(f,f',f'',\dots)$. We define now $$||f||_{k}=\sup\{||f||,||f'||,||f''||,\dots, ||f^{k)}||\}.$$

Prove that with the distance defined above $(||f-g||_k\\ \ $ $with\ f, g\in H^k(I))$ the space $H^k(I)$ is complete.

Okay, so now I have two questions.

First one: I've proved that $C_b(I)$, the space of all continuous bounded functions on $I$ is complete. Obviuosly, $H^k(I)\subset C_b(I)$, so if $C_b(I)$ is complete, $H^k(I)$ has to be complete to. What's wrong with my way of seeing this?

Obviously I must be doing something wrong, so the one stated above is not the correct answer. That leads me to think that maybe we can prove this exercise using or following a not so different path we followed while proving $C_b(I)$ is complete. But I get stuck all the time. Here's my try:

Let ${f_n}$ be a Cauchy sequence in $C_b(I)$. Fot every $x\in I$, we have that the sequence of real numbers $f_n(x)$ is a Cauchy sequence and it has to be convergent. Let's call its limit $f(x)$. We have then that $f_n(x)\to f(x)$. So, now we have to see that $f_n$ converges uniformly to $f(x)$.

Now I don't now how to move on from here. I'll appreciate any hint.

Thanks for your time.

Answer 1

It's not true that a subspace of a complete space is complete. For example, we have that $\mathbb{Q} \subset \mathbb{R}$, and we know $\mathbb{R}$ is complete, but $\mathbb{Q}$ is not.

Rather, we know that a closed subspace of a complete space is complete. However, this only holds if both the complete space and the subspace have the same norm. Using metric spaces instead of normed vector spaces (because I couldn't come up with a good example otherwise), in the discrete metric, $\mathbb{R}$ is complete and $\mathbb{Q}$ is closed in $\mathbb{R}$, but $\mathbb{Q}$ is not complete in the usual metric.

However, you have the advantage that $\|f\| \leq \|f\|_k$ for any $f \in H^k(\mathbb{R})$. Using this fact, your proof should start like this:

Let $(f_n)_n$ be a Cauchy sequence in $H^k(\mathbb{R})$. By the above inequality, $(f_n)_n$ is also a Cauchy sequence in $C_b(\mathbb{R})$, so by completeness, it converges to a function $f \in C_b(\mathbb{R})$.

Now you must show that $f$ is in fact infinitely differentiable, and all its derivatives are bounded. This should follow from the fact that this is true of all the $f_n$ and from the uniform convergence. I'll leave this for you to figure out!

P.S. Are you sure that $\|\cdot\|_k$ is in fact a norm? Are the norms of the derivatives guaranteed to be uniformly bounded? That is, can't the supremum be infinite, meaning the map isn't a norm?