Let $(X,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ be normed spaces, and $X\subset Y$. If each $f\in (X,\|\cdot\|_1)^\ast$ extends to a bounded linear functional in $(Y,\|\cdot\|_2)^\ast$ with same norm, then $(X,\|\cdot\|_1)$ is a subspace of $(Y,\|\cdot\|_2)$, i.e., for all $x\in X$, $\|x\|_1=\|x\|_2$.
I only proved that $\|x\|_1\leq\|x\|_2$ for all $x\in X$. I am in a difficult condition to prove $\|x\|_1\geq\|x\|_2$ for all $x\in X$.
Thanks a lot.
There is a counterexample: Let $Y=(R^2,\|\cdot\|_2)$ and $X=(R,\frac{1}{2}|\cdot|)$. For each linear functional $f=\langle a,\cdot\rangle$ on $X$, there is an extension $g=\langle a,\sqrt{3}a \rangle$ on $Y$ and $\|f\|=\|g\|$.