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.


Write \begin{align} |f(x)g(x)-f(y)g(y)|&\leqslant |f(x)\color{red}{g(x)}-f(y)\color{red}{g(x)}|+|\color{green}{f(y)}g(x)-\color{green}{f(y)}g(y)|\\ &\leqslant \color{red}{\lVert g\rVert_{\infty}}\lVert f\rVert_{\operatorname{LIP}}d(x,y)+\color{green}{\lVert f\rVert_{\infty}}\lVert g\rVert_{\operatorname{LIP}}d(x,y)\\ &\leqslant 2\max\{\lVert f\rVert_{\operatorname{BL}},\lVert g\rVert_{\operatorname{BL}}\}d(x,y). \end{align}


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