Concerning linear functionals on normed spaces Let $(X,||.||)$ be an normed vector space and $X^*$ its dual. In our lecture we recently stated that $$||x||=\sup_{\lambda \in X^*,||\lambda|| \leq 1} |\lambda(x)| $$ moreover , we stated that there is always such a $\lambda \in X^*$ with $||x||=\lambda(x)$, so the supremum is taken into place by some $\lambda$.
On the other hand for the definition of $||\lambda|| =\sup_{x \in X, ||x|| \leq 1}|\lambda(x)|$ there is not necessarily a $x \in X$ that takes shape of the supremum. 
Does this mean in some way, that the dual space $X^*$ of $X$ is greater or "richer" than $X$ in some way? I still have some struggles to get a good intuition of the dual space. 
 A: Concerning your question there are some quite useful lemmata and comments.


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*The "richness" of $X^*$ is guaranteed by the fact that for $x_1 \neq x_2 \in X$, $x = x_1 - x_2$ there is $x^* \in X^*$with $||x||=x^*(x)=x^*(x_1)-x^*(x_2) \neq 0$. So you can say that $x^*$ is able to tell both points appart. Any two distinct points in $X$ can be valued differently by some functional in $X^*$. 

*To understand how functionals work on $X$, it is necessary (as above mentioned) to think about the bidual-space $X^{**}$ of $X$: by defining the evaluation map  $$i:X \rightarrow X^{**}$$ by $$(i(x))(y):=y(x)$$ where $x \in X$ for every $y \in X^*$.  
Because $||i(x)||= ||x||$ holds, $i$ is an isometric (and clearly linear) map. So $i(X)$ is a subspace of $X^{**}$; By identifying any $x \in X$ with the function $x^{**}: X^* \rightarrow \mathbb{K}$ that plugs this very $x$ in every $x^* \in X^*$, you define an embedding of $X$ in $X^{**}$.  Additionally, if $i$ is surjective, you call X an reflexive space and $X \cong X^{**}$. 

*Concerning your question of norm-realization: apply the lemma you stated to $X^*$: for every $x^* \in X^*$ there exists a $x^{**}$ with $x^{**}(x^*)=||x^*||$. If $i$ is surjective, then there ex. a $x \in X$ with 
$$ ||x^*||= x^{**}(x^*)=i(x)(x^*)=x^*(x)$$ 
That means, reflexivity of $X$ guarantees that the supremum in the definition of $||x^*||$ is taken into place by some $x$.

