Let $A$ be the set of all continuous functions $f:[a,b] \to \mathbb{R}$ and let
$$d_2(f,g) = \left( \int_a^b |f(x)-g(x)|^2 dx\right)^{1/2}$$
Show that $d_2$ gives a metric.
\noindent sol) First show that Cauchy-Schwartz inequality. Let $f,g \in A$. Let $h(t) = \int_a^b (tf(x)+g(x))^2 dx$. Then $h(t) \geq 0$ for all $t \in \mathbb{R}$. $h(t) = (\int_a^b f(x)^2dx)t^2 + 2 \int_a^b f(x)g(x) dx t + \int_a^b g(x)^2 dx$. Discriminant $D$ of $h(t)=0$ is nonpositive, i.e. $D \leq 0$. It can be shown that $\int_a^b f(x)g(x) dx \leq \left(\int_a^b f(x)^2dx \int_a^b g(x)^2 dx \right)^{1/2}$ from the fact $D\leq 0$. $|f(x)-f(x)|=0$ for all $x \in [a,b]$, therefore, $d_2(f,f)=0$. $d_2(f,g)=0$ means that $|f(x)-g(x)|=0$ for all $x \in [a,b]$ due to the continuity of $f, g$ on $[a,b]$. Since $|f(x)-g(x)|=|g(x)-f(x)|$, $d_2(f,g)=d_2(g,f)$. $d_2(f,g) = \left(\int_a^b|f(x)-g(x)|^2dx \right)^{1/2} = \left(\int_a^b |f(x)-h(x)+h(x)-g(x)|^2dx \right)^{1/2} = \left(\int_a^b \left[ |f(x)-h(x)|^2+2(f(x)-h(x))(h(x)-g(x))+|h(x)-g(x)|^2 \right] dx \right)\leq \int_a^b |f(x)-h(x)|^2dx + \int_a^b |h(x)-g(x)|^2dx + 2 \left(\int_a^b (f(x)-h(x))^2dx \int_a^b (h(x)-g(x))^2 dx \right)$
$=\left(\left( \sqrt{\int_a^b(f(x)-h(x))^2dx} + \sqrt{\int_a^b(h(x)-g(x))^2dx}\right)^2\right)^{1/2}=d_2(f,h)+d_2(h,g)$. Therfore $d_2$ is a metric on $A$.
출처- 기하학1 (GIST-강현석교수님 강의) Homework 1 문제 중 일부