Let $\Omega$ be a measurable set in $R$. $X =L^{3}(\Omega)$ and $Z$ be a closed subspace of $X$. Let $x \in X$. Show that $\exists y \in Z$ such that $\lVert x-y\rVert=\inf_{z \in Z} ||x-z||=\delta$.
My attempt: By the definition of infimum, there exists a sequence $y_{n} \in Z$ such that $y_{n}$ such that $||x-y_n|| \rightarrow \delta$. We need to show that $y_{n}$ converges to a $y$ in Z.
Z is complete as its a closed subspace of a Banach space. If we can show that $y_{n}$ is a Cauchy sequence, then $y_{n}$ converges to a $y$ in $Z$.
$$\lVert y_{n}-y_{m}\rVert^2=2\lVert y_{n}-x\rVert^2+2\lVert x-y_{m}\rVert^2-\lVert y_{n}-2x+y_{m}\rVert^2$$ $$ \lVert y_{n}-y_{m}\rVert^2=2\lVert y_{n}-x\rVert^2+2\lVert x-y_{m}\rVert^2-4\left\lVert\frac{y_{n}+y_{m}}{2}-x\right\rVert^2$$
Taking limits as $ n,m \to \infty$, we get
$$\lVert y_{n}-y_{m}\lVert^2=4\delta^2-4\delta^2 \leq 0$$
Hence, $y_{n}$ is a Cauchy sequence.
Can someone verify this?
As pointed out in the comments, this proof is valid for $L^{2}$. I understand its wrong for $L^{3}$.
This proof works in any reflexive space $X$:
Since $(y_n)$ is norm bounded, Banach Alaoglu Theorem shows that $(y_n)$ has weakly convergent subsequence $(y_{n_k})$. Also, weak closure of a convex set is equal to the norm closure, so $Z$ is weakly closed. Hence, the weak limit $y$ of $(y_{n_k})$ belongs to $Z$. Now, if $f \in X^{*}$ with $\|f||\le 1$ then $|f(x-y)|=\lim |f(x-y_{n_k})| \leq \|x-y_{n_k}\|=\delta$. Taking supremum over $f$ we get $\|x-y\|\le \delta$. The reverse inequqality trivially holds.