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

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

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

First, we have \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 \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?