Given a Banach algebra $A$, define $\zeta: A \longrightarrow [0,\infty)$ by $\zeta(a) = \inf_{||c||=1}||ac||$. Prove that $|\zeta(a)-\zeta(b)| \leq ||a-b||$ for all $a,b \in A$.
So my first inclination was via the reverse triangle inequality, $$\big|||ac|| - ||bc|| \big| \leq ||ac-bc|| \implies \inf_{||c||=1} \big|||ac||-||bc|| \big| \leq \inf_{||c||=1} ||ac-bc|| \leq ||a-b||$$ so it suffices to show that $$\bigg|\inf_{||c||=1} ||ac|| - \inf_{||c||=1} ||bc|| \bigg| \leq \inf_{||c||=1} \big|||ac||-||bc|| \big| $$ which I tried showing by contradiction. Suppose that there exists $a,b \in A$ such that $|\zeta(a)-\zeta(b)| > \inf_{||c||=1}||ac-bc||$. Then there exists $c_1,c_2 \in A$ with $||c_k||=1$ $k=1,2$ such that: $$ \big||ac_1|| - ||bc_2|| \big| > ||ac-bc|| \implies ||ac-bc|| < ||ac_1-bc_2||$$ for all $c \in A$ with $||c||=1$, but I didn't see where else to go from here.
I also deduced pretty simply that $||ac|| - ||bc|| \leq ||ac-bc|| \implies \inf_{||c||=1}(||ac||) - ||bc|| \leq ||ac-bc||$ and that since $||bc|| \leq ||b||$ that: $$\inf_{||c||=1}(||ac||) - ||b|| \leq \inf_{||c||=1}(||ac||) - ||bc|| \implies \inf_{||c||=1}||ac|| - ||b|| \leq \inf_{||c||=1}||ac-bc||$$ but this doesn't let me take an infifum over all $||c||=1$ for $b$.
$\textbf{Something}$ tells me I'm missing something extremely obvious and vastly overcomplicating this simple problem. Any insight would help. Thanks!