# $\mathcal{C}^2[0,1]$ is a Banach Algebra

The following is problem 13 here.

Consider functions in $$\mathcal{C}^2[0,1]$$ and $$a,b>0$$. In this case, if we define: $$\lVert f \rVert:=\lVert f \rVert_\infty+ a \lVert f' \rVert_\infty +b \lVert f'' \rVert$$ This norm makes $$\mathcal{C}^2[0,1]$$ into a Banach algebra if and only if $$a^2\geq 2 b$$

I haven't been able to prove that being a Banach algebra implies $$a^2\geq 2 b$$. Here is the converse:

To prove that it is Banach whenever $$a,b>0$$ is straightforward. If $$f_n$$ is $$\lVert \cdot \rVert$$- Cauchy, we have uniform convergence of each derivative and we may write, $$g(x)=\lim f_n''(x)$$, $$h(x)=\lim f_n'(x)$$, $$f(x)=\lim f_n(x)$$. Because of uniform convergence of the sequence of continuous functions, $$g, h$$ and $$f$$ are continuous. Furthermore, because of uniform converge, we also have:

$$\int_a^x f_n'(u) du =f_n(x)-f_n(a) \quad \quad \int_a^x g(u) du=f(x)-f(a)\quad \quad f'(x)=g(x)$$

By similar reasoning, $$g'(x)=h(x)$$. So we are Banach as stated.

We need only verify product is compatible, namely:

$$\lVert fg \rVert\leq \lVert f\rVert \lVert g \rVert$$

Expanding things out yields:

$$\lVert f g\rVert_\infty+ a \lVert f'g+g'f \rVert_\infty +b \lVert f''g+2f'g'+g''f \rVert_\infty \leq (\lVert f \rVert_\infty+ a \lVert f' \rVert_\infty +b \lVert f'' \rVert)(\lVert g \rVert_\infty+ a \lVert g' \rVert_\infty +b \lVert g'' \rVert)$$

If we prove that in fact it holds that:

$$\lVert f\rVert_\infty \lvert g\rVert_\infty+ a \lVert f'\rVert_\infty \lVert g\rVert_\infty+a\lVert g'\rVert_\infty \lvert f \rVert_\infty +b \lVert f''\rVert \lVert g\rVert_\infty+2 b \lVert f '\rVert_\infty \lVert g'\rVert_\infty+b\lVert g''\rVert_\infty \lVert f \rVert\leq (\lVert f \rVert_\infty+ a \lVert f' \rVert_\infty +b \lVert f'' \rVert)(\lVert g \rVert_\infty+ a \lVert g' \rVert_\infty +b \lVert g'' \rVert)$$

Then we would be done. However, by rearraging factors we see this is equivalent to:

$$ab (\lVert f' \rVert_\infty \lVert g'' \rVert_\infty+\lVert f'' \rVert_\infty \lVert g' \rVert_\infty) +b^2 \lVert f'' \rVert_\infty \lVert g'' \rVert_\infty+(a^2-2b)\lVert f' \rVert_\infty \lVert g' \rVert_\infty\geq 0$$

This clearly holds if $$a^2-2b\geq 0$$ .

For the other implication, we need to find $$f,g\in C^2[0,1]$$ that do not satisfy $$\lVert fg \rVert\leq \lVert f\rVert \lVert g \rVert$$ if $$a^2-2b<0$$ any thoughts on what these functions should look like?

The problem actually states the condition $$2b\leq a^2$$, not $$b\leq a^2$$.
Consider $$f=g=x$$. Then $$\|f\|=1+a$$, $$\|f^2\|=1+2a+2b$$.
So we get that $$1+2a+2b=\|f^2\|\leq\|f\|^2=(1+a)^2=1+2a+a^2$$ which gives $$2b\leq a^2$$.
• I have corrected it just now. Thanks for pointing out the missing $2$. :) Feb 12 at 13:23