7. 设 `f(x)` 在 `[a, b]` 上有连续的 `n+1` 阶导数, 且 `f(a) = f'(a) = cdots` `= f^((n))(a) = 0`. 证明:
   1. `underset(a le x le b)(max) |f(x)|` `le (b-a)^n/(n!) int_a^b |f^((n+1))(x) dx|`;
   2. 对任意 `1 le p lt oo` 成立 `(int_a^b |f(x)|^p dx)^(1/p)` `le (b-a)^(n+1/p)/(n!(np+1)^(1/p)) int_a^b |f^((n+1))(x)| dx`;
   3. `underset(a le x le b)(max)|f(x)|` `le (b-a)^(n+1/2)/(n!sqrt(2n+1)) (int_a^b |f^((n+1))(x)|^2 dx)^(1/2)`.
   4. 对任意 `1 le p lt oo` 成立 `(int_a^b |f(x)|^p dx)^(1/p)` `le (2^(1/p) (b-a)^(n+1/2+1/p)) / (n!sqrt(2n+1)(2np+p+1)^(1/p)) (int_a^b |f^((n+1))(x)|^2 dx)^(1/2)`.
   由题目条件, `f(x)` 可以在 `a` 处展开成积分型余项的泰勒展式: `f(x) = 1/(n!) int_a^x (x-t)^n f^((n+1))(t) dt`, 于是
   1. `underset(a le x le b)(max)|f(x)|` `le (b-a)^n/(n!) int_a^b |f^((n+1))(t)| dt`.
   2. 对任意 `1 le p lt oo`, `int_a^b |f(x)|^p dx`
      `le 1/(n!)^p int_a^b (x-a)^(np) (int_a^b|f^((n+1))(t)|dt)^p dx`
      `= 1/(n!)^p (int_a^b (x-a)^(np) dx) (int_a^b |f^((n+1))(t)|dt)^p`
      `= 1/(n!)^p (b-a)^(np+1)/(np+1) (int_a^b |f^((n+1))(t)|dt)^p`. 两边同取 `1/p` 次方, 即得证.
   3. 利用柯西不等式. 对任意 `x in [a, b]`, `|f(x)|^2`
      `le 1/(n!)^2 (int_a^b (b-t)^n f^((n+1))(t) dt)^2`
      `le 1/(n!)^2 (int_a^b (b-t)^(2n) dt) (int_a^b |f^((n+1))(t)|^2 dt)`
      `= 1/(n!)^2 (b-a)^(2n+1)/(2n+1) int_a^b |f^((n+1))(t)|^2 dt`. 两边同时开平方根, 即得证.
   4. 兼用 (2) (3) 中的方法, `int_a^b |f(x)|^p dx`
      `= 1/(n!)^p int_a^b |int_a^x (x-t)^n f^((n+1))(t) dt|^p dx`
      `le 1/(n!)^p int_a^b (int_a^x (x-t)^(2n) dt)^(p/2)` `(int_a^b |f^((n+1))(t)|^2 dt)^(p/2) dx`
      `= 1/(n!)^p (int_a^b ((x-a)^(2n+1)/(2n+1))^(p/2) dx)` `(int_a^b |f^((n+1))(t)|^2 dt)^(p/2)`
      `= 1/(n!)^p (b-a)^(np+p/2+1)/((2n+1)^(p/2)(np+p/2+1))` `(int_a^b |f^((n+1))(t)|^2 dt)^(p/2)`
      `le 1/(n!)^p (2(b-a)^(np+p/2+1))/((2n+1)^(p/2)(2np+p+1))` `(int_a^b |f^((n+1))(t)|^2 dt)^(p/2)`. 两边同取 `1/p` 次方, 即得证.