Let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on a complex Hilbert space $E$.
On $\mathcal{L}(E)^2$, we have two equivalent norms:
\begin{eqnarray\*}
N\_1(A,B)
&=&\sup\left\{\sqrt{\|Ax\|^2+\|Bx\|^2},\;x\in E,\;\|x\|=1\;\right\},
\end{eqnarray\*}
and
$$N\_2(A,B)=\sqrt{\|A\|^2+\|B\|^2}.$$


Clearly, $N\_1(A,B)\leq N\_2(A,B)$.



> 
> In general $N\_1\neq N\_2$. I want to find sufficient conditions for equality.
> 
> 
> 



Sufficient condition for two norms to be equal
$\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand\ip[2]{\langle #1,#2\rangle}\newcommand\Span{\operatorname{span}}$As noted in a comment by [Yemon Choi](https://mathoverflow.net/questions/438068/sufficient-condition-for-two-norms-to-be-equal/438089#comment1129532_438068), the question is not well posed.


Apparently, the OP wanted to have a *good*, nontrivial sufficient condition, preferably close to necessity. Such a sufficient condition will be provided here, and the sufficient condition will also be necessary when $E$ is finite dimensional.


Let
\begin{equation}
 U:=A^\*A=\int\_{[0,\|A\|^2]}\la\,dP^A\_\la\quad\text{and}\quad 
V:=B^\*B=\int\_{[0,\|B\|^2]}\la\,dP^B\_\la \tag{0}\label{0}
\end{equation}
be the [spectral decompositions](https://en.wikipedia.org/wiki/Spectral_theorem#Spectral_subspaces_and_projection-valued_measures) of the self-adjoint bounded linear operators $A^\*A$ and $B^\*B$. (Note that the families $(P^A\_\la)$ and $(P^B\_\la)$ of orthoprojectors are uniquely determined by condition \eqref{0}.)


Let
$$Q^A\_\ep:=\int\_{[\|A\|^2-\ep,\|A\|^2]}dP^A\_\la, \quad Q^B\_\ep:=\int\_{[\|B\|^2-\ep,\|B\|^2]}dP^B\_\la,$$
$$E^A\_\ep:=Q^A\_\ep E,\quad 
E^B\_\ep:=Q^B\_\ep E. $$


Then the condition
\begin{equation}
 E^A\_\ep\cap E^B\_\ep\ne\{0\}\quad\forall\ep>0 \tag{1}\label{1}
\end{equation}
is sufficient for $N\_1(A,B)=N\_2(A,B)$.




---


Indeed, if \eqref{1} holds, then for each real $\ep>0$ there is a unit vector
$$u\_\ep\in E^A\_\ep\cap E^B\_\ep,$$
so that $Q^A\_\ep u\_\ep=u\_\ep=Q^B\_\ep u\_\ep$.
So,
$$\begin{aligned}
\|Au\_\ep\|^2&=\ip{A^\*Au\_\ep}{u\_\ep} \\ 
&=\int\_{[0,\|A\|^2]}\la\,d\ip{P^A\_\la u\_\ep}{u\_\ep} \\ 
&=\int\_{[0,\|A\|^2]}\la\,d\ip{P^A\_\la Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\ 
&=\int\_{[\|A\|^2-\ep,\|A\|^2]}\la\,d\ip{P^A\_\la Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\ 
&\ge(\|A\|^2-\ep)\int\_{[\|A\|^2-\ep,\|A\|^2]}d\ip{P^A\_\la Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\ 
&=(\|A\|^2-\ep)\ip{Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\ 
&=(\|A\|^2-\ep)\ip{u\_\ep}{u\_\ep}
=\|A\|^2-\ep. 
\end{aligned}$$
Similarly, $\|Bu\_\ep\|^2\ge\|B\|^2-\ep$. So,
$$N\_1(A,B)^2\ge\|Au\_\ep\|^2+\|Bu\_\ep\|^2
\ge\|A\|^2+\|B\|^2-2\ep=N\_2(A,B)^2-2\ep$$
for any real $\ep>0$.
Thus, $N\_1(A,B)\ge N\_2(A,B)\ge N\_1(A,B)$. $\quad\Box$




---


*Remark 1:* Condition \eqref{1} is necessary for $N\_1(A,B)=N\_2(A,B)$ when $E$ is finite dimensional. Indeed, suppose that $N\_1(A,B)=N\_2(A,B)$. Then for some sequence $(x\_n)$ on the unit sphere $S$ in $E$ we have
\begin{equation}
\|A\|^2+\|B\|^2\ge\|Ax\_n\|^2+\|Bx\_n\|^2\to\|A\|^2+\|B\|^2. 
\end{equation}
It follows that $\|Ax\_n\|^2\to\|A\|^2$ and $\|Bx\_n\|^2\to\|B\|^2$. The unit sphere $S$ in the finite-dimensional Hilbert space $E$ is compact. So, passing to a subsequence, without loss of generality we have $x\_n\to x$ for some $x\in S$. So, $\|Ax\|^2=\|A\|^2$ and $\|Bx\|^2=\|B\|^2$, which implies \eqref{1}. $\quad\Box$


*Remark 2:* Condition \eqref{1} is not in general necessary for $N\_1(A,B)=N\_2(A,B)$ when $E$ is infinite dimensional. E.g., let $E:=\ell^2$, with the standard basis $(e\_1,e\_2,\dots)$.


Let $U$ be the positive-semidefinite self-adjoint linear operator on $E$ whose matrix (in the standard basis) is the block-diagonal matrix with the $2\times2$ diagonal blocks
\begin{equation}
D\_j^U:=\frac j{j+1}\,P\_j^U,\quad\text{where}\quad 
P\_j^U:=\begin{bmatrix}1&0\\0&0 \end{bmatrix} 
\end{equation}
for integers $j\ge1$.
The factor $\frac j{j+1}$, strictly increasing to $1$ in $j$, was introduced to force gliding to $\infty$ in search of a (nonexistent) maximizer $x\in S$ of $\ip{Ux}x$.


Similarly, let $V$ be the positive-semidefinite self-adjoint linear operator on $E$ whose matrix (in the standard basis) is the block-diagonal matrix with the $2\times2$ diagonal blocks
\begin{equation}
D\_j^V:=\frac j{j+1}\,P\_j^V,\quad\text{where}\quad 
P\_j^V:=\frac1{j^2+1}\,\begin{bmatrix}j^2&j\\j&1 \end{bmatrix} 
\end{equation}
for integers $j\ge1$.
Note that $P\_j^U$ and $P\_j^V$ are orthoprojector matrices of rank $1$ each, with the respective column spaces $\Span\Big\{\begin{bmatrix}1\\0 \end{bmatrix}\Big\}$ and $\Span\Big\{\begin{bmatrix}j\\1 \end{bmatrix}\Big\}$.


In accordance with \eqref{0}, let $A:=\sqrt{U}$ and $B:=\sqrt{V}$. Then $\|A\|^2=\|U\|=1$ and $\|B\|^2=\|V\|=1$. Moreover, letting an integer $i$ go to $\infty$, we get $$N\_1(A,B)\ge\|Ae\_{2i}\|^2+\|Be\_{2i}\|^2=\ip{Ue\_{2i}}{e\_{2i}}+\ip{Ve\_{2i}}{e\_{2i}} \\ \to1+1=\|A\|^2+\|B\|^2=N\_2(A,B),$$
which implies that $N\_1(A,B)=N\_2(A,B)$.


However, for any integer $j\ge1$ and any $\ep\in(\frac1{j+2},\frac1{j+1}]$,
\begin{equation}
 E\_\ep^A=\Span\{e\_{2i}\colon i\ge j\}\quad\text{and}\quad
 E\_\ep^B=\Span\{e\_{2i}+\tfrac1{2i+1}\,e\_{2i+1}\colon i\ge j\},
\end{equation}
so that \eqref{1} fails to hold. $\quad\Box$




---


It appears that the following condition is necessary and sufficient for $N\_1(A,B)=N\_2(A,B)$:



> 
> There exist sequences $(a\_n)$ and $(b\_n)$ such that $a\_n-b\_n\to0$,
> $a\_n\in S\cap E\_{1/n}^A$ for all $n$, and $b\_n\in S\cap E\_{1/n}^B$ for all $n$.
> 
> 
> 


This conjecture is probably easy to prove, but ....


