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?


1 Answer 1


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 $2b\leq a^2$.

  • $\begingroup$ I have corrected it just now. Thanks for pointing out the missing $2$. :) $\endgroup$
    – Kadmos
    Feb 12 at 13:23

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