Proving an inequality involving $L_2(\Omega)$ norms of $L_2$ and $C^{\infty}$ functions

Question

Say $v\in L_2(\Omega)$, then need to show that there's a $w\in C_0^{\infty}(\Omega)$ such that $$\|v-w\|_{L_2(\Omega)}\leq\frac{1}{2}\|v\|_{L_2(\Omega)}$$.

So far I've done this: \begin{align*}\|v-w\|_{L_2(\Omega)}^2=&\int_{\Omega}(v-w)^2\\ =&\int_{\Omega}(v^2+w^2-2vw)\\ =&\int_{\Omega}v^2+\int_{\Omega}w^2-2\int_{\Omega}vw\\ =&\|v\|_{L_2}^2+\|w\|_{L_2}^2-2(v,w)_{L_2}\\ \leq&\|v\|_{L_2}^2+\|w\|_{L_2}^2\end{align*}

If this is correct way, what to do after this? If not, how should I proceed?

This is a statement (w/o any explanation) from the Finite Element text (p. 146) by Braess.

Answer 1

As @Jake28 mentioned, this just follows from the density of $C^\infty_c(\Omega)$ in $L^2(\Omega)$. That is for any $v\in L^2(\Omega)$ and any $\varepsilon>0$ there exists $w_\varepsilon\in C^\infty_c(\Omega)$ such that $$\|v-w_\varepsilon\|_{L^2(\Omega)}\le\varepsilon.$$

This is simply a case of choosing the correct $\varepsilon$, i.e., here we want $\varepsilon = \frac{1}{2}\|v\|_{L^2(\Omega)}$.