counter example of minimum principle in incomplete inner product space minimum principle:
Minimum principle in Hilbert space
I want to construct a counter example when $H$ is a inner product space but not a Hilbert space. Can we find a closed convex subset of $H$ such that there is no minimum point( or more than one) in it? 
 A: Let $H = L^2([0,2])$ with the $L^2$ inner product and consider the linear functional $\varphi \colon H \rightarrow \mathbb{C}$ given by 
$$ \varphi(f) := \int_0^{1} f(x) \, dx. $$
Note that $\varphi$ is continuous as
$$ |\varphi(f)| = \left| \int_0^1 f(x) \, dx \right| \leq \int_0^1 |f(x)| \cdot 1 \, dx \leq \left( \int_0^1 |f(x)|^2 \right)^{1/2} \left( \int_0^1 1^2 \, dx \right)^{1/2} \leq \\
\left( \int_0^2 |f(x)|^2 \right)^{1/2} = ||f|| $$
and so the affine subspace
$$ V := \varphi^{-1}(1) = \left \{ f \in L^2([0,2]) \, | \, \int_0^1 f(x) \, dx = 1 \right \} $$
is a closed and convex subspace of $H$. The norm estimate of $\varphi$ shows that if $f \in V$ then $||f|| \geq 1$ and the minimum is attained uniquely for
$$ g(x) = \begin{cases} 1 & 0 \leq x \leq 1, \\ 0 & 1 < x \leq 2 \end{cases} $$
which is not (equal a.e to) a continuous function.

Now, consider $V \cap C([0,2])$ as an affine subspace of the space $C([0,2])$, endowed with the $L^2$ norm. The set $V \cap C([0,2])$ is convex and closed (as a subset of $C([0,2])$) and $\inf_{f \in V \cap C([0,2])} ||f|| = 1$ but it doesn't have an element of norm $1$ since such an element would also be an continuous function of minimal norm of $V$ inside $H$ which is impossible.
