Let $\displaystyle (X,d)$ metric space. Let $\displaystyle Y=\mathcal B\mathcal C(X,\mathbb R)$ be the space of every continuous and bounded functions from $\displaystyle X$ to $\displaystyle \mathbb R.$

Prove that $\displaystyle h(f,g)=\underset{x\in X}{\mathop{\sup }}\,\left| f-g \right|$ is a metric on $\displaystyle Y.$

Well obviously $\displaystyle h(f,g)\ge0,$ if $\displaystyle f=g\implies h(f,g)=0,$ besides it's $\displaystyle h(f,g)=h(g,f),$ so the only thing we need to prove is that $\displaystyle h(f,g)\le h(f,j)+h(j,g).$

First, we have $\displaystyle \left| f-g \right|=\left| f-j+j-g \right|\le \left| f-j \right|+\left| j-g \right|,$ but it's easy to see that every component of the inequality is bounded, so it follows that $\displaystyle \underset{x\in X}{\mathop{\sup }}\,\left| f-g \right|\le \underset{x\in X}{\mathop{\sup }}\,\left| f-j \right|+\underset{x\in X}{\mathop{\sup }}\,\left| j-g \right|.$

Is this correct?