# Equivalence of norms on the space of smooth functions

Let $E, F$ be Banach spaces, $A$ be an open set in $E$ and $C^2(A,F)$ be the space of all functions $f:A\to F,$ which are twice continuously differentiable and bounded with all derivatives. The question is when following two norms in $C^2(A,F)$ are equivalent: $$\|f\|_{1}=\sup_{x\in A}\sum^2_{k=0}\|f^{(k)}(x)\|, \ \|f\|_{2}=\sup_{x\in A}(\|f(x)\|+\|f^{(2)}(x)\|).$$ In the case $A=E$ they are equivalent. One can prove it in the following way. To bound $\|f^{(1)}(x)h\|$ consider a line $g(t)=f(x+th)$ through $x$ in the direction $h$ and use inequality $$\sup_{t\in \mathbb{R}}\|g^{(1)}(t)\|\leq\sqrt{2\sup_{t\in \mathbb{R}}\|g(t)\|\sup_{t\in \mathbb{R}}\|g^{(2)}(t)\|}.$$ The case when $A$ is an open ball is unknown to me. Of course, one can try to consider not lines but segments. But the problem is the length of segment can't be bounded from below and inequalitites I know can't be applied.

-

We can reduce to the case $F=\mathbb R$ by considering all compositions $\varphi\circ f$ with $\varphi$ ranging over unit-norm functionals on $F$.

Let $A$ be the open unit ball in $E$. Given $x\in A$ and direction $v\in E$ (a unit vector), we would like to estimate the directional derivative $f_v(x)$ in terms of $M=\sup_A(\|f\|, \|f\,''\|)$. Instead of restricting to a line, let us restrict $f$ to a 2-dimensional subspace $P$ that contains the line (and the origin). The advantage is that $D:=A\cap P$ is a 2-dimensional unit disk: the size of section does not depend on $x$ or $v$.

The directional derivative $f_v:D\to \mathbb R$ is itself differentiable, and using the 2nd derivative bound we conclude that the oscillation of $f_v$ on $A$ (namely, $\sup_A f_v-\inf_A f_v$) is bounded by $2M$. Suppose $\sup_A f_v>5M$. Then $f_v>3M$ everywhere in $A$. Integrating this inequality along the line through the origin in direction $v$, we find that the oscillation of $f$ on this line is at least $6M$, a contradiction. Therefore, $\sup_A f_v\le 5M$. Same argument shows $\inf_A f_v\ge -5M$.

-