 Solution

The bound in part (b) and domination $l\le p$ give $l(y)\le L\Vert y\Vert$. Applying the same bound to $-y$, and using linearity, gives $-l(y)=l(-y)\le L\Vert y\Vert$. ConsequentlyA bounded linear functional is continuous, so $l$ belongs to the continuous dual space $E^{\prime }$. The domination by a sublinear functional alone was algebraic; the local Lipschitz continuity supplies the needed boundedness.