Let $M,N$ be two normed vector spaces with norms $\|\cdot\|_M,\|\cdot\|_N$, respectively. $A_1,A_2$ are two sets in the space $M\oplus N$, whose norm is defined by \begin{equation} \|(f,g)\| = \sqrt{\|f\|_M^2+\|g\|_N^2}\,,\quad f\in M, g\in N\,. \end{equation} We define the distance \begin{equation} d(A_1,A_2) = \max\left\{\sup_{(f_1,g_1)\in A_1}\inf_{(f_2,g_2) \in A_2}\frac{\|(f_1,g_1)-(f_2,g_2)\|}{\|(f_1,g_1)\|},\sup_{(f_2,g_2)\in A_2}\inf_{(f_1,g_1)\in A_1}\frac{\|(f_1,g_1)-(f_2,g_2)\|}{\|(f_2,g_2)\|}\right\}\,. \end{equation}
Let $T_1, T_2$ be two bounded linear operators from $M$ to $N$. I want to show the following:
If $A_1$ and $A_2$ are the graphs of $T_1$ and $T_2$, i.e., $A_i = \{(f,T_if):f \in M\}, i = 1,2$, then \begin{equation} \frac{\|T_1-T_2\|_{M\rightarrow N}}{\sqrt{1+\|T_1\|_{M\rightarrow N}^2}\sqrt{1+\|T_2\|_{M\rightarrow N}^2}}\leq d(A_1,A_2) \leq \|T_1-T_2\|_{M\rightarrow N} \end{equation} where the norm $\|\cdot\|_{M\rightarrow N}$ is the operator norm, i.e., $\|T\|_{M\rightarrow N} = \sup_{f\in M} \frac{\|Tf\|_{N}}{\|f\|_M}$
Any hint will be appreciated. Thanks.