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?

Edit:
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 $O(\log r)$?

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?

Edit:
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 $O(\ln r)$?

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?

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?

Edit:
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 $O(\log r)$?