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Let $T_{n\times n}$ be a triangular truncation matrix, i.e. $$T_{i,j}=\begin{cases}1 & i\ge j\\ 0 & i<j \end{cases}$$ It is known that for arbitrary $A_{n\times n}$ $$\|T\circ A\|\le\frac{\ln n}{\pi}\|A\|$$ where $\circ$ is the hadamard product.

For example, a proof of above was given in this paper

If $A$ is of rank $r<n$, a simple inequality yields: $$\|T\circ A\|\le\|A\|_F\le \sqrt r\|A\|$$ where $\|\cdot\|_F$ is the Frobenius norm

Therefore for rank deficient $A$, the bound can be smaller.
I was wondering if this $\sqrt r$ is sharp enough, is there an example to show $$\|T\circ A\|/\|A\|\to c\sqrt r$$ For A of varying size but fixed rank.
Or is it possible to improve $\sqrt r$ to something even smaller, like the logarithm in general case?

There're some misunderstanding about my question, I re-organized it as follow:
Let $A_{n\times n}$ of rank $r$, where $$\frac{\ln n}{\pi}>\sqrt r$$ Let $n$ grow in someway but keep $r$ fixed, is it possible to construct an example to show $$\frac{\|T\circ A\|}{\|A\|}\sim O(\sqrt r)$$ or to show the ratio is actually $O(\ln r)$?

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It is perhaps useful to point out that what you and Bhatia (in the linked paper) refer to as the "Frobenius norm" is called the Hilbert-Schmidt norm by many (including me). – Christian Remling Jul 27 '14 at 20:59
Also, at the risk of being pedantic, we can add some precision to the question: let $f(r)=\sup \| T\circ A\|/\|A\|$, where the $\sup$ is over all (non-zero) $n\times n$ matrices of rank $r$, with arbitrary $n$. Then we would like to know if $\liminf r^{-1/2}f(r)>0$. (In your version, if taken at face value, any expression can go on the RHS since I can always absorb everything by the constant $c$; we need to send $r\to\infty$ also eventually.) – Christian Remling Jul 27 '14 at 21:45
up vote 5 down vote accepted

The ratio is of order $O(\ln r)$. This follows from the fact that the triangular truncation is bounded on the Schatten class $S^p$ (=the operators $A$ on $\ell^2$ such that $\|A\|_p:= (Tr (A^*A)^{p/2})^{1/p}$ is finite) and has norm $O(p)$ as $p \to \infty$. Indeed, for $A \in M_n$ of rank $r$ and operator norm $1$, then for all $p \geq 2$ $$ \|T \circ A\| \leq \|T \circ A\|_p \leq C p \|A\|_p \leq C p r^{1/p}.$$ I suffices to take $p = \ln r$.

The fact that the triangular truncation has norm $O(p)$ on $S^p$ is well-known. One reference that might contain a proof of this fact (or contain a reference) is the paper E. Davies, Lipschitz continuity of functions of operators in the Schatten classes, J. London Math. Soc. 37 (1988), 148–157. This is also related to the fact that the Hilbert transform is completely bounded on $L^p$ with norm $O(p)$ for $p \geq 2$.

Edit I had a look at the paper by Davies that I mention above, and indeed he uses that the triangular truncation $T$ is bounded on $S^p$ with norm $O(p)$. For this fact, he refers to a book by Krein and Gohberg Theory and Applications of Volterra operators in Hilbert spaces, where they attribute the result to Krein and Gohberg, On the theory of triangular representations of nonselfadjoint operators (1961). I am not sure I understand exactly how their result formally implies that $T$ is bounded, but here is a very simple proof that is probably their proof (and that is very close to the standard proof of the boundedness of the Hilbert transform on $L^p$).

Denote by $C_p$ the max of the norms of $T$ and $(1-T)$ on $S^p$ (actually it is not hard to see that $T$ and $(1-T)$ have the same norms, but this is of no use here). Then $C_2=1$, and if we prove that $C_{2p}\leq 2 C_p$ we have by interpolation that $C_p \leq p$ for all $p\geq 2$. To prove that $C_{2p} \leq C_p$, remark that for every $A$, $$ (TA)^* (TA) = T( (TA)^* A) + (1-T)( A^* TA )$$ Hence $$ \|TA\|_{2p}^2 = \| (TA)^* (TA) \|_p \leq \|T( (TA)^* A)\|_p + \|(1-T)( A^* TA )\|_p,$$ and each of the two terms is less than $C_p \|A^* TA\|_p \leq C_p \|A\|_{2p}\|TA\|_{2p}$ by Hölder's inequality. If we divide both sides by $\|TA\|_{2p}$ and take the supremum over $A$ we get the result.

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Thanks! This looks great – galaas Sep 28 '14 at 12:43

Is this square root bound such a good one, compared to the general case ?

Maybe for some fixed $n$ and small $r$, but in general with $r = \Theta(n)$ (viz. $r = n-1$), your $\sqrt{r}$-bound is weaker than the $\log n$ one : $\sqrt{r} \gg \log n$.

So this bound cannot be exact... otherwise $\|T\circ\cdot\| \gg \sqrt{n}$, and we know that it is at worst in $\log n$.

Or am I misunderstanding the question ?

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I believe that the idea is to let n grow but fix r. Then $\sqrt{r}$ is smaller than $\log(n)$, and the question is whether or not we can do better than $\sqrt{r}$ (e.g., $\log(r)$). – Nathaniel Johnston Jul 27 '14 at 20:23
Johnston's comment is of correct understanding, I re-organized my question to make it clearer – galaas Jul 28 '14 at 19:34

Fix $r \geq 1$ and let $A_n$ be the $n \times n$ identity matrix with the bottom $n-r$ rows replaced with rows of zeros. Then $||A_n|| = 1$ for all $n$ and $||T_n \circ A_n|| = ||A_n||$ so we have that $\frac{||T_n \circ A_n ||}{||A_n||} = 1 \leq \sqrt{r}$ for all $n$.

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The question is asking for the best possible upper bound in terms of r, so what is needed is either an example that gets the ratio much larger than 1 (i.e., close to $\sqrt{r}$) or a proof of a better upper bound (e.g., something like $\log(r)$). – Nathaniel Johnston Jul 28 '14 at 13:41
Johnston's comment is of correct understanding, I re-organized my question to make it clearer – galaas Jul 28 '14 at 19:33

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