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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?

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  • $\begingroup$ 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
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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

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

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    $\begingroup$ This is a very clever "simplification"! $\endgroup$ – Iosif Pinelis Dec 6 '19 at 14:31
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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 $C \ge 0$.

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    $\begingroup$ 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
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    $\begingroup$ @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

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