Operator on polynomials, antiderivativeWe are given a linear map: R [ X ] ∋ p → q ∈ R [ X ] , q ′ = p , q ( 0 ) = 0 and two norms on R [ X ] : | | p | | ∞ = sup t ∈ [ 0 , 1 ] | p ( t ) | , | | p | | 1 = ∫ 0 1 | p ( t ) | d t.I want to check whether the map is continuous in these norms.When it comes to the first norm, I get: | | q | | ∞ = sup t ∈ [ 0 , 1 ] | q ( t ) | By Lagrange mean value theorem we have that there exists a ∈ [ 0 , 1 ] such that q ′ ( a ) = q ( t ) − q ( 0 ) t , so q ( t ) ≤ sup [ 0 , 1 ] | q ′ ( t ) | .And | | p | | ∞ = sup t ∈ [ 0 , 1 ] | p ( t ) | = sup t ∈ [ 0 , 1 ] | q ′ ( t ) | ≥ | | q | | Is that correct so far?

Question

Operator on polynomials, antiderivativeWe are given a linear map:      R    [  X  ]  ∋  p  →  q  ∈      R    [  X  ]  ,            q    ′    =  p  ,        q  (  0  )  =  0 and two norms on       R    [  X  ] :       |        |    p      |              |              ∞        =      sup          t      ∈      [      0      ,      1      ]            |    p  (  t  )      |  ,       |        |    p      |              |        1    =      ∫    0    1        |    p  (  t  )      |    d  t.I want to check whether the map is continuous in these norms.When it comes to the first norm, I get:      |        |    q      |              |              ∞        =      sup          t      ∈      [      0      ,      1      ]            |    q  (  t  )      |  By Lagrange mean value theorem we have that there exists   a  ∈  [  0  ,  1  ] such that       q    ′    (  a  )  =            q      (      t      )      −      q      (      0      )        t  , so   q  (  t  )  ≤      sup          [      0      ,      1      ]            |        q    ′    (  t  )      |  .And       |        |    p      |              |              ∞        =      sup          t      ∈      [      0      ,      1      ]            |    p  (  t  )      |    =      sup          t      ∈      [      0      ,      1      ]            |        q    ′    (  t  )      |    ≥      |        |    q      |        |  Is that correct so far?

Punktatsp · Accepted Answer

Step 1Your work is correct, but in my opinion this makes things more clear:  q  (  x  )  =      ∫    0    x    p  (  t  )    d  t  .Then, for the   ‖        ‖    ∞   norm      |    q  (  x  )      |    ≤      ∫    0    x        |    p  (  t  )      |      d  t  ≤  x    ‖  p      ‖    ∞    ≤  ‖  p      ‖    ∞    .Step 2Fot the   ‖        ‖    1   norm       ∫    0    1        |    q  (  x  )      |      d  x  =      ∫    0    1              |            ∫    0    x    p  (  t  )    d  t            |          d  x  ≤      ∫    0    1        ∫    0    1        |    p  (  t  )      |      d  t    d  x  =  ‖  p      ‖    1    .

Operator on polynomials, antiderivative We are given a linear map: <mrow class="MJX-TeXAtom-OR

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