# Strictly convex norm on an infinite-dimensional Hilbert space

Question: Consider the Hilbert space $H=\ell^2(\mathbb{Z})$. Let ${\rm L}(H)$ be the set of all linear operators on $H$ onto itself. Find a norm $N$ and a domain $DN\subset {\rm L}(H)$ for $N$ satisfying simultaneously the following three conditions: (i) $DN$ is convex; (ii) $DN \ni I$ (identity operator) and (iii) $N$ is strictly convex. That is, $N( (1 - t)x + ty ) < (1 - t) N(x) + tN(y)$ for any $t\in (0,1)$ and $x,y\in DN$ with $x\neq y$.

In affirmative case, what is the largest $DN$ we can take?

• Maybe, $N((1-t)x+ty)<(1-t)N(x)+tN(y)$ unless $x,y$ are collinear? Commented Jun 2, 2016 at 5:32
• Also, do you consider all linear operators or only bounded? And how should this norm be connected to action of $L(H)$ on $H$? Without such assumptions we may look at $L(H)$ as simply on a linear space. Commented Jun 2, 2016 at 5:34
• Your title seems a bit misleading. Perhaps you mean to say "strictly convex norm on a space of operators on Hilbert space"? Commented Jun 2, 2016 at 17:16
• The classical way of getting an equivalent strictly convex norm on a Banach space $X$ is to find an injective bounded linear operator $T$ from $X$ into some strictly convex space $Y$ and use $\|x\| +\Tx\|$. When $X=L(\ell_2)$, you can do this with $Y=\ell_2$. Commented Jun 3, 2016 at 18:20

I think, we may take even $DN=L(H)$ by setting $$N(B)=\sup_{\|x\|=1} \|Bx\|+\sum 2^{-n}\|Bx_n\|,$$ where $x_1,\dots$ is a dense subset of a unit ball in $H$. The second term is majorated by the first, so it is equivalent and is not hard to prove that it is strictly convex.
For an example, take $DN=$ the sum of $\mathbb R I$ and of the set of Hilbert-Schmidt operators $(Au)_k:=\sum_j a_{j,k}u_j$ with $\sum|a_{j,k}|^2<\infty$, and $N(cI+A)^2=c^2+\sum|a_{j,k}|^2$. It makes $DN$ itself a Hilbert space, the norm on which is (as always) "strictly convex" for non collinear $x$ and $y$.