Is it true that for all symmetric n × n matrix, ( a i j ), such that for x ∈ { ± 1 } n ∑ i j a i j x i x j ≤ 1there exists a universal constant such that ∑ i j a i j x i y j ≤ Cfor all x , y ∈ { ± 1 } n . I tried to use the polarization identity ⟨ A x , y ⟩ = ⟨ A u , u ⟩ − ⟨ A v , v ⟩where u = ( x + y ) / 2 and v = ( x − y ) / 2. However, as x and y vary over ± 1 vectors, u and v can be vectors in { ± 1 , 0 }.

Question

Is it true that for all symmetric   n  ×  n matrix,   (      a          i      j        ), such that for   x  ∈  {  ±  1      }    n        ∑          i      j            a          i      j            x    i        x    j    ≤  1there exists a universal constant such that      ∑          i      j            a          i      j            x    i        y    j    ≤  Cfor all   x  ,  y  ∈  {  ±  1      }    n   . I tried to use the polarization identity  ⟨  A  x  ,  y  ⟩  =  ⟨  A  u  ,  u  ⟩  −  ⟨  A  v  ,  v  ⟩where   u  =  (  x  +  y  )      /    2 and   v  =  (  x  −  y  )      /    2. However, as   x and   y vary over   ±  1 vectors,   u and   v can be vectors in   {  ±  1  ,  0  }.

Jamir Rojas · Accepted Answer

No, e.g.       x    T        (                            n                                              −          n                      )    x  ≡  0 for every   {  −  1  ,  1  }-vector   x, but   (  1  ,  1  )      (                            n                                              −          n                      )        (                            1                                      −          1                      )    =  2  n is unbounded.

Is it true that for all symmetric n × n matrix, ( a <

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