Let C ≥ 2 and L &gt; 0 be fixed. Let n ∈ { 1 , 2 , … }. Does there exist a polynomial g of degree n such that 0 &lt; g ( 0 ) &lt; g ( L ) , g ′ ( x ) &gt; ( g ( L ) − g ( x ) ) C , for all x ∈ [ 0 , L ]holds at least for n big enough?

Question

Let   C  ≥  2 and   L  &amp;gt;  0 be fixed. Let   n  ∈  {  1  ,  2  ,  …  }. Does there exist a polynomial g of degree n such that  0  &amp;lt;  g  (  0  )  &amp;lt;  g  (  L  )  ,        g    ′    (  x  )  &amp;gt;  (  g  (  L  )  −  g  (  x  )  )  C  ,        for all   x  ∈  [  0  ,  L  ]holds at least for n big enough?

rioolpijpgp · Accepted Answer

Step 1For any fixed   L  ,  C  &amp;gt;  0 , pick n sufficient large so that   L  &amp;lt;      n    C   . Now define the polynomial,  p  (  x  )  :=  −            (      −              C                  2          n                    x      +              1        2            )              2      n        +  1Note that p(x) satisfies the points   (  −  n      /    C  ,  0  ) ,   (  0  ,  1  −  (  1      /    2      )          2      n        ) ,   (  n      /    C  ,  1  ) ,   (  2  n      /    C  ,  1  −  (  1      /    2      )          2      n        ) , which implies that p(x) has a vertex at   n      /    C (the absolute maximum of p), and that p(x) is increasing from   (  −  ∞  ,  n      /    C  ] (moreover, p(x) has positive derivative and is in fact concave down on that same region); additionally we see that   0  &amp;lt;  p  (  x  )  &amp;lt;  1 for   x  ∈  [  0  ,  L  ] . Thus   0  &amp;lt;  −      C          2      n        x  +      1    2    &amp;lt;  11 for   x  ∈  [  0  ,  L  ] . Additionally, we clearly see that   0  &amp;lt;  p  (  0  )  &amp;lt;  p  (  L  )  &amp;lt;  1 . So for any   x  ∈  [  0  ,  L  ] we have                              p          ′                (        x        )                    =                    −        2        n                              (            −                          C                              2                n                                      x            +                          1              2                        )                                2            n            −            1                          ⋅                              −            C                                2            n                                                    =                    C                              (            −                          C                              2                n                                      x            +                          1              2                        )                                2            n            −            1                                                    &amp;gt;                    C                              (            −                          C                              2                n                                      x            +                          1              2                        )                                2            n                                                    &amp;gt;                    C                  (                                    (              −                              C                                  2                  n                                            x              +                              1                2                            )                                      2              n                                −                                    (              −                              C                                  2                  n                                            L              +                              1                2                            )                                      2              n                                )                                          =                    C                  (          p          (          L          )          −          p          (          x          )          )

Let C ≥ 2 and L > 0 be fixed. Let n ∈<!--

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