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?

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    $\begingroup$ $K$ is not closed, your last paragraph proves it. $a(n)$ is a sequence in $K$ which converges to $0 \notin K$. $\endgroup$ – Nate Eldredge Oct 26 '16 at 4:24
  • $\begingroup$ @NateEldredge indeed :) $\endgroup$ – Takanashi Oct 26 '16 at 4:38

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.

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  • $\begingroup$ How do you confirm that $C([0,2])$ is closed $\endgroup$ – Takanashi Oct 26 '16 at 21:42
  • $\begingroup$ @Takanashi: $C([0,2])$ is not closed inside $L^2([0,2])$. However, since $V$ is closed inside $L^2([0,2])$ then $V \cap C([0,2])$ is closed inside $C([0,2])$. $\endgroup$ – levap Oct 26 '16 at 22:13
  • $\begingroup$ Wy this Is true? Is there any topological theory related to your inference? $\endgroup$ – Takanashi Oct 26 '16 at 23:19
  • $\begingroup$ @Takanashi: Yes. If $X$ is a topological space and $A \subseteq X$ is a closed set then if $V \subseteq X$ is any subset of $X$ then $A \cap V$ is a closed subset of $V$ (where $V$ has the induced topology). This is because the complement of $A \cap V$ in $V$ is $(X \setminus A) \cap V$ which is open by the definition of the induced topology. $\endgroup$ – levap Oct 27 '16 at 4:00

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