积分绝对值不等式

已知函数$f(x) \in C^{(1)}[a,b] $，求证：

(1)对于任意的$x \in [a,b]$，有
\[ |f(x)| \le \left| \frac{1}{b-a} \int_{a}^{b} f(x) \mathrm{d} x\right|+ \int_{a}^{b} |f'(x)| \mathrm{d} x \]
(2)
\[ \left |f \left ( \frac{a+b}{2} \right ) \right| \le \left|\frac{1}{b-a} \int_{a}^{b} f(x) \mathrm{d} x\right|+ \frac{1}{2} \int_{a}^{b} |f'(x)| \mathrm{d} x \]

①不妨设 $f\left( \xi \right) = \mathop {\max}\limits_{a \leqslant x \leqslant b} \left| {f\left( x\right)} \right|$，$f\left( \eta\right) = \frac{1}{{b - a}}\int_a^b {f\left( x \right){\text{d}}x} $

\[\int\limits_a^b{\left|{f'\left( x \right)}\right|{\text{d}}x} \geqslant\int\limits_\eta^\xi {\left|{f'\left(x\right)}\right|{\text{d}}x}\geqslant\left|{f\left(\xi\right)-f\left(\eta\right)}\right|\geqslant\left|{f\left(\xi\right)}\right|-\left|{f\left(\eta\right)}\right|\geqslant\left|{f\left(x\right)}\right|-\left|{f\left(\eta\right)}\right|\]

②设 $m = \frac{{a + b}}{2}$，由第一题，
\[\left| {f\left( m \right)} \right| \leqslant \left| {\int\limits_a^b {f\left( x \right){\text{d}}x} } \right| + \int\limits_a^m {f'\left( x \right){\text{d}}x} \]
\[\left| {f\left( m \right)} \right| \leqslant \left| {\int\limits_a^b {f\left( x \right){\text{d}}x} } \right| + \int\limits_m^b {f'\left( x \right){\text{d}}x} \]

相加，\[\left| {f\left( m \right)} \right| \leqslant \left| {\int\limits_a^b {f\left( x \right){\text{d}}x} } \right| + \frac{1}{2}\int\limits_a^b {f'\left( x \right){\text{d}}x} \]