Let us regard the $n\times n$ matrices as operators on the $n$-dimensional $\ell_p$ space; that is, we consider them as linear operators $\ell_p^n\to \ell_p^n$. When $p=2$, $M_n$ is a C*-algebra and we have

$0 \leqslant A \leqslant B \implies \|A\|\leqslant \|B\|$.

Here $\|A\|$ denotes the operator norm of a map $A\colon \ell_2^n\to \ell_2^n$. What about other $p\in [1,\infty]$?

Fix $p\in [1,\infty]$. Is it true that there exists $K>0$ such that for every $n$ and for all $A,B\colon \ell_p^n\to \ell_p^n$ with $0\leqslant A\leqslant B$ (meaning that $A$ and $B$ are self-adjoint and non-negative semi-definite) we have $$\|A\|_{\ell_p^n\to\ell_p^n}\leqslant K\|B\|_{\ell_p^n\to\ell_p^n}?$$

My feeling is that it should be true for $p\in (1,\infty)$. For $p=1$ or $p=\infty$ there is an easy counter-example with $K=1$. Take

$$A=\left[\begin{smallmatrix}2&1\\1& \tfrac{1}{2}\end{smallmatrix}\right],\;\;B = \left[\begin{smallmatrix}\tfrac{5}{2}&0\\0& \tfrac{5}{2}\end{smallmatrix}\right]. $$

Then $0\leqslant A\leqslant B$ yet for $p\in \{1,\infty\}$ we have $\|A\|_{\ell_p^2\to\ell_p^2} = 3$ whereas $\|B\|_{\ell_p^2\to\ell_p^2}=\tfrac{5}{2}$.

In the language of this thread: Monotone matrix norms, I ask whether the operator $\ell_p$-norms are monotone, that is, if we can take $K=1$. user147215 cleverly shows that this is not the case when $p\neq 2$.

Possible approach: It is not inconceivable that using some Riesz–Thorin-type argument we could show that the operator $\ell_p$-norms are indeed monotone for $p$ is some neighbourhood of 2.


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  • $\begingroup$ Please share the counterexample for $p=1,\infty$, if you don't mind. $\endgroup$ – user147194 May 1 '14 at 18:38
  • $\begingroup$ Since you ask if there is a $K=K_p$ independent of $n$; my instinct (which I haven't checked) is to find some $n$ and some $0\leq A \leq I_n$ with $\Vert A\Vert_{p\to p} > 1$ for some $p\neq 2$, and then look at tensor powers of $A$. $\endgroup$ – Yemon Choi May 1 '14 at 23:07
  • $\begingroup$ The answer is no, Tomek. Use a Kashin decomposition of $L_p^n$ to see that a random orthogonal projection has norm at least $Cn^{|1/p-1/2|}$. $\endgroup$ – Bill Johnson May 2 '14 at 1:15

The answer is no, Tomek. Use a Kashin decomposition of $L_p^n$, $1\le p < 2$, to see that there are orthogonal projections whose norms as operators on $L_p^n$ are of order $Cn^{|1/p-1/2|}$. (Kashin proved that for $1\le p < 2$ there is an orthogonal decomposition $A+B$ of $n$-space s.t. if $x \in A \cup B$, then $\|x\|_p \le \|x\|_2 \le C\|x\|_p$, where I am using the uniform probability measure on $\{1,\dots,n\}$ rather than counting measure to define the norms.)

The case $p>2$ follows by duality.

B. S. Kashin, Sections of some finite-dimensional bodies and classes of smooth functions. Izv. Acad. Nauk SSSR41 (1997), 334--351.

  • $\begingroup$ Thank you. I didn't know Kashin's decomposition. Actually this suggests a question of whether the converse holds: if this condition is satisfied, must the induced operator norms come from Euclidean spaces? I think I can prove that the answer is no if the finite-dimensional spaces $E_n$ (which now generalise $\ell_p^n$ in the statement) are ranges of initial projections of a symmetric basis in a space which contains $\ell_1^n$ or $\ell_\infty^n$ uniformly complemented. $\endgroup$ – Tomek Kania May 2 '14 at 13:00
  • 1
    $\begingroup$ Sounds likely. The condition is equivalent to saying that there is a constant $C$ s.t. every orthogonal basis has unconditional constant at most $C$. I bet that this implies that the norm is $f(C)$ equivalent to the Euclidean norm. $\endgroup$ – Bill Johnson May 2 '14 at 13:10
  • $\begingroup$ I'm glad to hear that; looks like a nice exercise! $\endgroup$ – Tomek Kania May 2 '14 at 13:14
  • $\begingroup$ Let me point out that for $K=1$ and any $n$-dimensional space the claim follows from the Kakutani-Sobczyk-Bohnenblust theorem characterising isometrically Hilbert spaces as those Banach spaces whose all 2-dimensional subspaces are 1-complemented. $\endgroup$ – Tomek Kania Jun 5 '14 at 23:12

The original version of the question, which asked for $\|A\|_{\ell_p^n\to\ell_p^n}\leqslant K\|B\|_{\ell_p^n\to\ell_p^n}$ with $K = K(p)$, was more interesting (and I don't have an answer to that one yet).

But if you insist on $K=1$, then the inequality fails for all $p\ne 2$. Let $n=3$, $$A=\begin{pmatrix} 2/3 & -1/3 & -1/3 \\ -1/3 & 2/3 & -1/3 \\ -1/3 & -1/3 & 2/3 \end{pmatrix},\qquad B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ That is, $B=I$ and $A=I-P$ where $P$ is the orthogonal projection onto the line $x_1=x_2=x_3$. What $A$ does is subtract the mean of $x_i$ from each coordinate.

Subtracting the mean does not increase the $\ell_2$ norm of a vector, but it may increase $\ell_p$ for any other $p$. I will show that $\|A\|_{\ell_p^n\to\ell_p^n}>1= \|B\|_{\ell_p^n\to\ell_p^n}$.

Fix $p\in (1,\infty)\setminus \{2\}$ and let $x = (2^{1/(p-1)},-1,-1)^T$. For $p\in (1,\infty)$ the function $$\phi(t) = \sum_i |x_i-t|^p$$ is strictly convex, and attains its minimum when $\phi'(t)=0$, namely at the point with $$ \sum_i |x_i-t|^{p-1}\operatorname{sign}(x_i-t)=0 \tag{1}$$ For $x$ as above, (1) is satisfied with $t=0$. Therefore, $\phi(t)>\|x\|_p^p$ for all $t\ne 0$. Applying $A$ amounts to replacing each $x_i$ with $x_i-t$, where $t= (2^{1/(p-1)}-2)/3$. When $p\ne 2$, we have $t \ne 0$, hence $\|Ax\|_p> \|x\|_p$.


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