# Equivalence of operators

let $$T$$ and $$S$$ be positive definite (thus self-adjoint) operators on a Hilbert space.

I am wondering whether we have equivalence of operators

$$c(T+S) \le \sqrt{T^2+S^2} \le C(T+S)$$ for some appropriate $$c$$ and $$C$$.

This is somewhat motivated by the equivalence of all $$p$$ norms on a Hilbert space.

The first inequality is trivial

since $$T^2\le T^2+S^2$$ and $$S^2 \le T^2+S^2$$ we conclude since the square root is operator monotone that $$\frac{T+S}{2} \le \sqrt{T^2+S^2}.$$

However, I do not find it very obvious whether

$$\sqrt{T^2+S^2} \le C(T+S)$$

is possible?

• Welcome to MathOverflow! Do you want $C$ to be independent of $T$ and $S$? – Jochen Glueck Dec 5 '19 at 21:40
• @JochenGlueck thank you for your comment. That is precisely what I want. – van Dyke Dec 5 '19 at 21:40

There is no real $$C>0$$ such that $$\sqrt{T^2+S^2} \le C(T+S) \tag{1}$$ holds for all positive definite (self-adjoint) operators on a Hilbert space of any dimension $$\ge2$$.

Indeed, take any any real $$C>0$$ and identify $$T$$ and $$S$$ with the $$2\times2$$ matrices $$T:=\left( \begin{array}{cc} 1 & 0 \\ 0 & s^5 \\ \end{array} \right),\quad S:=\left( \begin{array}{cc} \dfrac{s^2}{4} & \dfrac{s^3}{8} \\ \dfrac{s^3}{8} & \dfrac{s^6}{64}+\dfrac{s^4}{16} \\ \end{array} \right),$$ where $$s:=1/C.$$ If (1) holds for some real $$C>0$$, then it holds for any large enough $$C>0$$, because $$T+S\ge0$$. Now, letting $$C\to\infty$$ (so that $$s\downarrow0$$), we have $$\det\big(C(T+S)-\sqrt{T^2+S^2}\,\big)=-\frac{s^2}{16}+O\left(s^3\right)<0$$ for all large enough $$C>0$$.

Thus indeed, there is no real $$C>0$$ such that (1) holds for all positive definite operators on a Hilbert space of any dimension $$\ge2$$.

Here is an image of the Mathematica notebook with the relevant calculations:

A slightly simplified version of Iosif Pinelis' counterexample: for any $$n\ge1$$ let

$$S:=\left[ \begin {array}{cc} 1& \sqrt{n-1}\\ \sqrt{n-1}&n\end {array} \right],\qquad T:=\left[ \begin {array}{cc} 1&- \sqrt{n-1}\\ - \sqrt{n-1}&n\end {array} \right]$$

Then $$S+T=2\, {\rm diag}\big( 1,\, n\big)$$ while $$S^2+T^2 =2\, {\rm diag}\big( n,\, n^2+n-1\big),$$
so $$\sqrt{S^2+T^2 } =\sqrt{2}\, {\rm diag}\big( \sqrt{n }\,,\, \sqrt{n^2+n-1}\big),$$

and comparing the $$(1,1)$$ entries, we have that $$C$$ must be at least $$\sqrt\frac{{n}}{2}$$ in order that $$(S^2+T^2)^{1/2}\le C(S+T).$$

• This is a very clever "simplification"! – Iosif Pinelis Dec 6 '19 at 14:31

An additional remark (too long for a comment) that might be of interest:

Given the finite-dimensional counterexamples in the answers by Iosif Pinelis and Pietro Majer it seems worthwhile to note that we can use those examples to construct a counterexample in infinite dimension which is "stronger" in the sense that, for fixed $$T$$ and $$S$$, there does not exist any $$C \ge 0$$ the satisfies the required inequality, but "weaker" in the sense that the operators involved are only positive semi-definite:

Example. There exist positive semi-definite operators $$T$$ and $$S$$ on the separable complex Hilbert space such that $$\sqrt{T^2 + S^2} \le C (S+T) \qquad (*)$$ does not hold for any $$C \ge 0$$.

Indeed, for each $$n \in \mathbb{N}$$ there exist, by the other answers, operators $$T_n,S_n$$ on $$\mathbb{C}^2$$ such that $$\sqrt{T_n^2 + S_n^2} \not\le n (S_n+T_n)$$. Since $$(*)$$ is invariant under multiplication of both $$T$$ and $$S$$ with the same positive number, we may assume that $$\|T_n\| \le 1$$ and $$\|S_n\| \le 1$$ for each $$n$$.

Let us then consider the operators $$T = \oplus_{n \in \mathbb{N}} T_n$$ and $$S = \oplus_{n \in \mathbb{N}} S_n$$ on the Hilbert space $$\ell^2(\mathbb{N}; \mathbb{C}^2)$$; it follows that those two operators do not satisfy $$(*)$$ for any $$C \ge 0$$.

• In this example, the operators T and S are both not invertibile. I suspect that if at least one of them is invertible, the original inequality would be true. – Pietro Majer Dec 10 '19 at 4:34
• @PietroMajer: Yes indeed; if $A,B$ are positive semi-definite operators on a Hilbert space and $B$ is invertible, then there exist real numbers $c_1,c_2 > 0$ such that $c_1A \le \operatorname{id} \le c_2B$. (Ah, but now I see that the OP required $T$ and $C$ to be positive definite rather than positive semi-definite; I'll edit my answer and note this explicitly). – Jochen Glueck Dec 10 '19 at 9:22