Existence of continous functional over infinite dimensional space without global minimum The Weierstrass theorem says that any continuous function over compact set attains it extreme value within the set. Since compact means closed and bounded only for finite dimensional spaces, my guess is that the theorem does not necessarily hold for infinite dimensional spaces. For example the unit sphere in an infinite dimensional space is not compact even though it is closed and bounded (from converse of Reisz's Lemma, which I do not profess to fully understand - last paragraph of question).
So to show the theorem wrong, I am trying to construct an explicit example of a continuous functional $J: A \subseteq V \to \mathbb{R}$ where $V$ is a function space and $A$ is a closed and bounded subset of $V$ such that the global minimum of $J$ over $A$ does not exist. 
To this end, are there any limitations on the norm or metric I choose to imbue the problem with? As in are there norms, which if I choose, the example I am trying to construct will always fail? Also does anyone know of an example which matches what I am trying to do?
I don't have a strong background in analysis and I am trying to learn what I can for an optimization course that I am taking. So I am not sure where to start off either. If someone could give some pointers, sketches, or references it would be really great. 

Let $V= C[0, 1]$, with $J$ the sup norm. Let $A$ consist of all $f$ in $V$ such that $\|f\|\leq 2$ and
$$\int_{0}^{\frac{1}{2}}f(s)ds-\int_{\frac{1}{2}}^{1}f(s)ds=1.$$
Then $A$ is a closed and bounded subset of $V$ which contains no element of minimum norm, that is the minimum of $J$ over $A$ does not exist.
In fact 
$$
1 = \left|\int_0^{\frac{1}{2}} f(s) \, ds - \int_{\frac{1}{2}}^1 f(s) \, ds\right|  \leq \int_0^{\frac{1}{2}} \|f\| \, ds +\int_{\frac{1}{2}}^1 \|f\| \, ds 
= \|f\|.
$$
Moreover, for $n\geq 3$ let $f_n: [0,1] \to \mathbb{R}$ be the function whose graph is the union of the following line segments from $(0,1)$ to $(\frac{1}{2}, 1)$, from $(\frac{1}{2},1)$ to $(\frac{1}{2} + \frac{1}{n}, \frac{1+n}{1-n})$, and from $(\frac{1}{2} + \frac{1}{n}, \frac{1+n}{1-n})$ to $(1, \frac{1+n}{1-n})$.  It is easy to verify that $\|f_n\| = \frac{n+1}{n-1}$ and $f_n \in A$. Since $\|f_n\|\to 1$, it follows that
$\inf \{\|f\|: f \in A\} =1$.
However there is no $f \in A$ such that $\|f\| = 1$.
In fact, assume that $f$ in $A\subset C[0,1]$ satisfies $\|f\| = 1$.  Then 
$$1 = \int_0^{\frac{1}{2}} f(s) \, ds - \int_{\frac{1}{2}}^1 f(s) \, ds$$
implies that $f(s)=1$ in $[0,1/2)$ and $f(s)=-1$ in $(1/2,1]$ (otherwise the LHS is less than $1$) and $f$ can not be continuous in $[0,1]$.
For a more detailed proof see here: A vector without minimum norm in a Banach space
