matrix inequality upper bound $\|A(B+C) +D(B-C)\|/2$?

Question

Suppose $A,B,C,D$ are matrices with operator norm at most 1. Then is it true that $\|A(B+C) +D(B-C)\|/2$ is at most $1$ (where $\|\cdot \|$ is the operator norm of a matrix, where by operator norm, I mean largest eigenvalue of the matrix)? I have tried many examples and it seems true, but I am not able to prove something in general.

Answer 1

$A=I$, $B=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, $C=D=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. All of these have an induced two norm of one.

Then $A(B+C)+D(B-C) = \begin{bmatrix} 2 & 0 \\ 2 & 0 \end{bmatrix}$ which has an induced norm of $2 \sqrt{2} > 2$.

octave:1> A = [ 1 0 ; 0 1 ] ;
octave:2> B = [ 1 0 ; 0 -1 ] ;
octave:3> C = D = [ 0 -1 ; 1 0] ;
octave:4> 
octave:4> norm(A)
ans =  1
octave:5> norm(B)
ans =  1
octave:6> norm(C)
ans =  1
octave:7> norm(D)
ans =  1
octave:8> 
octave:8> S = A*(B+C)+D*(B-C)
S =

   2   0
   2   0

octave:9> norm(S)
ans =  2.8284
octave:10> 
octave:10> sqrt(eig(S'*S))
ans =

   0.00000
   2.82843