Does a pointwisely convergent sequence of quadratic functions pointwisely converges to a quadratic one?Let { f j } j = 1 ∞ be a sequence of functions f j : R n → R quadratically parameterized as f j ( x ) = x T P j x for a symmetric matrix P j ∈ R n × n . Let f j pointwisely converges to a function f.Then, can we say that f is also quadratic so that f ( x ) = x T P x for some symmetric P ∈ R n × n ?I was particularly interested in the case: fj is monotonically increasing and bounded by a quadratic function g ( x ) = x T Q x: f 1 ( x ) ≤ ⋯ ≤ f j ( x ) ≤ f j + 1 ( x ) ≤ ⋯ ≤ g ( x ) &lt; ∞ ∀ x ∈ R n .In that case, by MCT, f j → f pointwisely for some f, and we also have P j → P for some symmetric P ∈ R n × n , by MCT again, due to P 1 ≤ ⋯ ≤ P j ≤ P j + 1 ≤ ⋯ ≤ Q .In this case, f ( x ) = x T P x is true since x T P j x → x T P x for all x ∈ R n . Moreover, if f ( x ) = x T P x is true, then it is trivially continuous, so that the convergence is uniform on any compact Ω ⊂ R n by Dini's theorem.However, I can't see whether f ( x ) = x T P x is true or not in general case without monotonicity.

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Does a pointwisely convergent sequence of quadratic functions pointwisely converges to a quadratic one?Let   {      f    j        }          j      =      1        ∞   be a sequence of functions       f    j    :            R        n    →      R   quadratically parameterized as       f    j    (  x  )  =      x    T        P    j    x for a symmetric matrix       P    j    ∈            R              n      ×      n      . Let       f    j   pointwisely converges to a function f.Then, can we say that f is also quadratic so that   f  (  x  )  =      x    T    P  x for some symmetric   P  ∈            R              n      ×      n      ?I was particularly interested in the case: fj is monotonically increasing and bounded by a quadratic function   g  (  x  )  =      x    T    Q  x:      f    1    (  x  )  ≤  ⋯  ≤      f    j    (  x  )  ≤      f          j      +      1        (  x  )  ≤  ⋯  ≤  g  (  x  )  &amp;lt;  ∞        ∀  x  ∈            R        n    .In that case, by MCT,       f    j    →  f pointwisely for some f, and we also have       P    j    →  P for some symmetric   P  ∈            R              n      ×      n      , by MCT again, due to      P    1    ≤  ⋯  ≤      P    j    ≤      P          j      +      1        ≤  ⋯  ≤  Q  .In this case,   f  (  x  )  =      x    T    P  x is true since       x    T        P    j    x  →      x    T    P  x for all   x  ∈            R        n  . Moreover, if   f  (  x  )  =      x    T    P  x is true, then it is trivially continuous, so that the convergence is uniform on any compact   Ω  ⊂            R        n   by Dini&#039;s theorem.However, I can&#039;t see whether   f  (  x  )  =      x    T    P  x is true or not in general case without monotonicity.

Faith Wise · Accepted Answer

Step 1You may use a polarization identity which gives us      B    j    (  x  ,  y  )  =      x    T        P    j    y  =      1    2        (          f      j        (    x    +    y    )    −          f      j        (    x    −    y    )    )  Step 2This determines a unique symmetric bilinear form in x,y. The hypothesis is that       f    j   converges pontwise so from the above identity:  B  (  x  ,  y  )  =      lim          j      →      ∞            B    j    (  x  ,  y  )exists for all x,y. One checks it is bilinear in x and y, so has the form       x    T    P  y. One recovers then that   f  (  x  )  =      lim    j        f    j    (  x  )  =  B  (  x  ,  x  )  =      x    T    P  x.

Rihanna Bentley · Accepted Answer

Step 1The result for   n  =  1 is straightforward. Suppose we know the result for             R        n    , and that       Q    j    (  x  ,  y  ) is a sequence of quadratics on             R              n      +      1       that converges pointwise on             R              n      +      1        .. Here   x  =  (      x    1    ,  …  ,      x    n    )  . We can write      Q    j    (  x  ,  y  )  =      R    j    (  x  )  +      a    j    y  +  y  (      b          j      1            x    1    +  ⋯  +      b          j      n            x    n    )  +      c    j        y    2    .Here       R    j   is a quadratic on             R        n    , and the coefficients of       Q    j   are those of       R    j   along with the coefficients involving y indicated above.Step 2Looking at (x,0), the induction hypothesis shows the all sequences of coefficients of       R    j    (  x  ) converge, hence       R    j    (  x  ) converges pointwise on             R        n    .. Thus       Q    j    −      R    j   converges pointwise on             R              n      +      1        . If we now look at the variables   (  0  ,      x    1    ,  …  ,      x    n    ,  y  )  , the induction hypotheses tells us all remaining coefficent sequences converge, except for       b          j      1        . But now this last coefficient sequence is forced to come along for the ride, and we have the result.

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