Let $X$ be a separable Banach space and $T_1:D(T_1) \subset X \rightarrow X$ and $T_2:D(T_2) \subset X \rightarrow X$ two closed operators with $D(T_2)\subset D(T_1)$ and $D(T_2^*) \subset D(T_1^*).$

We say that $T_1$ is relatively $T_2$ bounded if $D(T_2) \subset D(T_1)$ and for all $x \in D(T_2)$

$$\left\lVert T_1x \right\rVert \le \alpha \left\lVert T_2 x \right\rVert + \beta \left\lVert x \right\rVert.$$

I am interested in the following question:

Are there sufficient conditions (on the operators or the space for example $X$ being reflexive) such that $T_1^*$ being relatively $T_2^*$ bounded implies that $T_1$ is $T_2$ bounded with the same relative bound?

Let $X$ be a separable Banach space and $T_1:D(T_1) \subset X \rightarrow X$ and $T_2:D(T_2) \subset X \rightarrow X$ two closed operators with $D(T_2)\subset D(T_1)$ and $D(T_2^*) \subset D(T_1^*).$

We say that $T_1$ is relatively $T_2$ bounded if $D(T_2) \subset D(T_1)$ and for all $x \in D(T_2)$

$$\left\lVert T_1x \right\rVert \le \alpha \left\lVert T_2 x \right\rVert + \beta \left\lVert x \right\rVert.$$

I am interested in the following question:

Are there sufficient conditions such that $T_1^*$ being relatively $T_2^*$ bounded implies that $T_1$ is $T_2$ bounded with the same relative bound?

What about the converse: Does $T_1$ being relatively $T_2$ bounded imply that $T_1^*$ is relatively $T_2^*$ bounded under the above conditions?