Solution
= Solution
The bound in part (b) and domination $l\leq p$ give $l(y)\leq L\|y\|$. Applying the same bound to $-y$, and using linearity, gives $-l(y)=l(-y)\leq L\|y\|$. Consequently
$$
\boxed{|l(y)|\leq L\|y\|,\qquad\|l\|\leq L.}
$$
A <bounded linear functional> is continuous, so $l$ belongs to the <continuous dual space> $E'$. The domination by a <sublinear functional> alone was algebraic; the local <Lipschitz continuity> supplies the needed boundedness.