The Theorem says "$H^1(a,b)$ is compactly embedded in $L^2(a,b)$".
In the proof, it is written that the continuity follows directly from $$\Vert \cdot \Vert_{H^1}^2=\Vert \cdot \Vert_{L^2}^2+\vert \cdot\vert_{H^1}^2$$
But the definition of an embedding being continuous is that there exists $\alpha > 0$ such that $$\Vert j(v) \Vert_{L^2} \leq \alpha \Vert v\Vert_{H^1}$$ for all $v \in H^1(a,b)$.
What obviousness am I missing?
Best Answer
We have $j(v)=v$ and then $$\|j(v)\|^2_{L^2}= \|v\|_{L^2}^2 \le \|v\|_{L^2}^2 + |v|_{H_1}^2 =\|v\|_{H^1}^2. $$ Hence taking square roots, $\|j(v)\|_{L^2}\le \|v\|_{H^1}$ so you can choose $\alpha=1$.