# Inequalities for bounded lipschitz functions

Suppose $$X$$ is a metric space with metric $$d$$. Define $$\lVert f(x)\rVert_\infty = \sup_x |f(x)|$$ and $$\lVert f(x)\rVert_{\mathrm{LIP}}=\sup\left\{\frac{|f(x)-f(y)|}{d(x,y}:x\neq y\right\}$$, let $$\lVert f\rVert|_{BL}=\max(\lVert f\rVert_\infty,\lVert f\rVert_{\mathrm{LIP}})$$ and $$\mathrm{BL}(X)=\{f:\lVert f\rVert_{\mathrm{BL}}<\infty\}$$, i.e. $$\mathrm{BL}(X)$$ contains all the functions $$f$$ such that $$f$$ is bounded and Lipschitz on $$X$$. I am trying to show for $$f,g\in \mathrm{BL}(X)$$, $$\lVert fg\rVert_{\mathrm{BL}}\le 2\lVert f\rVert_{\mathrm{BL}} \cdot\lVert g\rVert_{\mathrm{BL}}$$.

My approach so far: there is no problem to show $$\lVert fg\rVert_\infty\le \lVert f\rVert_\infty\lVert g\rVert_\infty$$. Just follow the definition, we have

$$\lVert fg\rVert_\infty=\sup_x |f(x)g(x)|\le \sup_x |f(x)|\cdot|g(x)|\le \sup_x |f(x)|\sup_x |g(x)|=\lVert f\rVert_\infty\lVert g\rVert_\infty.$$

For $$\lVert fg\rVert_{\mathrm{LIP}}\le 2\lVert f\rVert_{\mathrm{LIP}}\cdot\lVert g\rVert_{\mathrm{LIP}}$$ ( I guess the $$2$$ must be here since I don't use it for the infinity norm), I don't see how can I achieve this. I can prove $$fg\in \mathrm{BL}(X)$$, but this seems lead nowhere.