# Sufficient condition for two norms to be equal

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.

• When does $\sup (f + g)$ equal $\sup f + \sup g$? Jan 8 at 11:39
• Neither of them is homogenous. Did you forget square roots? Jan 8 at 12:13
• @JochenWengenroth Yes I'm sorry. I corrected them Jan 8 at 13:11
• Here is a sufficient condition: A=B=0. Jan 9 at 0:18
• The operators $A$ and $B$ are not zero Jan 9 at 4:29

$$\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand\ip{\langle #1,#2\rangle}\newcommand\Span{\operatorname{span}}$$As noted in a comment by Yemon Choi, 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 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 ....

• I think this should also be a necessary condition, isn’t it? Jan 8 at 20:41
• You seem to have lost the square on lambda in the first formula. Or you should specify the projectors' definitios. Jan 9 at 0:23
• @PietroMajer : This condition is of course necessary if $E$ is finite dimensional. However, it is now shown that it is not necessary in general. Jan 9 at 2:25
• @tsnao : I do not think anything is lost here. However, it is now added "that the families $(P^A_\lambda)$ and $(P^B_\lambda)$ of orthoprojectors are uniquely determined by condition (0)". Jan 9 at 2:29