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?

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

enter image description here


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 $$ (S^2+T^2)^{1/2}\le C(S+T).$$

  • 1
    $\begingroup$ This is a very clever "simplification"! $\endgroup$ – 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 $C \ge 0$.

  • 1
    $\begingroup$ 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. $\endgroup$ – Pietro Majer Dec 10 '19 at 4:34
  • 1
    $\begingroup$ @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). $\endgroup$ – Jochen Glueck Dec 10 '19 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.