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. 
 A: Though the above answer is satisfactory, I think I have a simpler and less contrived example; please feel free to comment on any holes in the argument. Let $ V = \mathcal{C}\left( {\left[ {0,1} \right],\mathbb{R}} \right)$ and let  $$A = \left\{ {f \in V \mid f\left( 0 \right) = 0,f\left( 1 \right) = 1} \right\} \subset V$$ Let $J = \left\| f \right\| = \int_0^1 {f\left( x \right)dx} $. Then $A$ is bounded above by one under this norm. Note that the functions $ \left\{ {x,{x^2},{x^3}...} \right\} \in A$ and have smaller and smaller values for the functional $J$. However, for any function of the form $ x^n$ the function $x^{n+1}$ will have a smaller value of the functional. In fact the smallest possible value of $J$ under the constraints is achieved by 
$$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {0\quad x \ne 1} \\ 
  {1\quad x = 1} 
\end{array}\quad ,\quad x \in \left[ {0,1} \right]} \right.$$
which does not belong to $V$ (that is it is not sequentially compact, and the Weierstrass theorem does not apply). Hence there is no minimum for $J$ over $A$. 
I have left out the part where I need to show $A$ is closed. The idea is to work with the definition of a closed set and the complement of a open set. For example a subset of the complement of $A$ can be defined as, 
$$ B = \left\{ {f \in V|f\left( 0 \right) = 0,f\left( 1 \right) \ne 1} \right\} \subset {A^C} \subset V $$
Then for any $ \bar f \in B$ arbitrarily close to 1 at  $\bar f\left( 1 \right)$ there exists a $ \hat f \in B$ that is closer. Thus $B$ is open. Similar argument can be made by relaxing the other end point or even both at the same time and so $A^C$ is open. Therefore its complement $A$ must be closed. 
A: 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
A: For the last answer, the function $$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {0\quad x \ne 1} \\ 
  {1\quad x = 1} 
\end{array}\quad ,\quad x \in \left[ {0,1} \right]} \right.$$ is the limit point of $x^n$. And it is not in the set so A should not be closed right ? (I apologize for putting it here since I am new to the website and can not comment.) I came up with the following argument. Please comment if you find any loopholes/errors.
Consider $V= L^p(E)$ where  $E$ is of finite measure. Then $L^q(E)\subset L^p(E)$ for any $p<q<\infty$. Let $A = L^q(E)$ equipped with $L^q$ norm. Since $L^q$ is complete, we have closedness. Boundedness is obtained by definition. Consider the functional defined on $A$ as follows: 
$$J = \frac{1}{\int_0^1 |f(x)|^2 dx}$$
Then $\inf J =0$ and is not achieved $\forall f$ $\in$ $A$.
