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For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate continuity and linearity make both maps bounded linear operators. For each fixed $y$,The Uniform boundedness principle therefore gives $C<\infty$ such that $\Vert L_x\Vert \le C$ whenever $\Vert x\Vert \le 1$. Scaling yieldsFor completeness, the relevant Uniform boundedness principle follows from the Baire category theorem as follows. For a pointwise bounded family $\mathcal{T}$ of bounded linear operators on a Banach space, the closed sets $E_m=\{y:{\sup }_{T\in \mathcal{T}}\Vert Ty\Vert \le m\}$ cover the space. One $E_m$ contains an open ball $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\Vert h\Vert <r$, gives ${\sup }_T\Vert Th\Vert \le 2m$. Applying this to $h=(r/2)u$, $\Vert u\Vert \le 1$, bounds every operator norm by $4m/r$.

Define the new norm byIt is a norm because $L_{x}e=x$, so $L_x=0$ implies $x=0$. The triangle inequality and scalar homogeneity follow from those of the operator norm. Moreover,so it is equivalent to the original norm and remains complete. By associativity, $L_{xy}=L_x{L}_y$, and the submultiplicativity of the operator norm provesThis is renorming a separately continuous Banach algebra. It even gives $\Vert e{\Vert }_{\ast }=1$. If the algebra is the zero space, the conclusion is immediate separately.