继续积分绝对值不等式

已知函数$f(x) \in C^{(1)}[a,b]$，且$f(a)=0$，求证：
\[ \int_{a}^{b} |f(x)f'(x)| \mathrm{d} x \le \frac{b-a}{2} \int_{a}^{b} |f'(x)|^2 \mathrm{d} x\]

1# 海盗船长


证明：
设$ F(x)=|f(x)|-|f(a)|=\int_{a}^{x}{|f(t)|dt} $
由此\[ |f(x)|=F(x) \]
\[ |f'(x)|=F'(x)\]
\[ \Rightarrow \int_{a}^{b}{F(x)F'(x)dx}=\frac{1}{2}F^{2}(b) =\frac{1}{2}(\int_{a}^{b}{|f'(t)|dt})^{2}\]
Now,Using Cauchy-Schwarz
\[ \frac{1}{2}(\int_{a}^{b}{|f'(t)|dt})^{2}\leq \frac{b-a}{2}\int_{a}^{b}{(f'(x))^{2}dx} \]
Done!