Restriction and boundedness of a map

Solution 1:

I assume that by $\hat{A}$ you mean the unitization. I will denote it by $\tilde{A}$ instead and I will establish a bit of different notation. Fix $x\in\tilde{A}$ and define $L_x:\tilde{A}\to\tilde{A}$ by $L_x(y)=xy$. This is obviously a linear mapping.

Note that if $y\in A$, then $L_x(y)$ in $A$; indeed, the inclusion $A\subset\tilde{A}$ is by identifying $A$ with the set $\{(a,0):a\in A\}$, so, if $x=(a_x,\lambda_x)$ and $y\equiv(y,0)\in A$, we have $$L_x(y)=(a_x,\lambda_x)\cdot(y,0)=(a_xy+\lambda_xy+0\cdot a_x,\lambda_x\cdot0)=(a_xy+\lambda_xy,0)\in A.$$ This shows that the restriction of $L_x$ on $A$ does indeed take values in $A$.

I will also assume that by $\|\cdot\|_\infty$ of a bounded map you mean the usual operator norm. Note that for $y\in A$ one has $\|L_x(y)\|=\|(a_xy+\lambda_xy,0)\|=\|a_xy+\lambda_xy\|\le(\|a_x\|+|\lambda_x|)\cdot\|y\|=\|x\|_1\cdot\|y\|$ for all $y\in A$, so $\|L_x\|\le\|x\|_1$.