Let g : R + → R , defined as follows : g ( x ) = x 1 + x , it's concave ( since g ″ ( x ) ≤ 0).Now we define the sup-norm on C ∞ [ 0 , 1 ], ψ ( f ) = sup x ∈ [ 0 , 1 ] | f ( x ) | .prove that g o ψ is concave.

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Let   g  :            R      +        →      R  , defined as follows :   g  (  x  )  =      x          1      +      x      , it&#039;s concave ( since       g    ″    (  x  )  ≤  0).Now we define the sup-norm on       C          ∞        [  0  ,  1  ],   ψ  (  f  )  =      sup          x      ∈      [      0      ,      1      ]            |    f  (  x  )      |  .prove that   g  o  ψ is concave.

wlanauee · Accepted Answer

Let   f  =  1 and   g  =  −  1. If   g  ∘  ψ is concave then   g  (  ψ  (            1      +      (      −      1      )        2    )  ≥            g      (      ψ      (      1      )      +      g      (      ψ      (      −      1      )        2  . The left side is   0. but   g  (  ψ  (  1  )  )  =  g  (  ψ  (  −  1  )  )  =      1    2   so the right side is   1.

Let g:R+->R, defined as follows : g(x)=x/1+x, it's concave ( since g′′(x)<=0).Now we define the sup-norm on Coo[0,1], psi (f)=sup_(x in[0,1])|f(x)|. Prove that go psi is concave.

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