Meaning of Riesz representations in a layman's term? I have read Riesz representation theorem quite a lot of times. While understanding the mathematical meaning, that is, for each element $x$ in a Hilbert space $X$, there will be a linear and bounded functional $f_x$ in $X$ that when acts on any $y\in X$, is inner product of $x$ with $y$ and vice versa.
I am struggling to have a deeper understanding of this. What are Riesz representers? Can these be viewed as geometrical objects? Why are these so important? How can someone explain importance of this theorem to a layman?
Any help and correction will be appreciated, thanks. 
 A: I don't know if this answers your question, I will intepret "layman" as "someone who knows linear algebra". I think a nice intuition for Riesz Representation theorem is thinking of it as the infinite dimensional equivalent of transposing a vector. In finite dimension (say 3) the isomorphism between a vector space and its dual is obvious, if you write a vector $x$ in some basis
$$ x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$
it is obvious what is the functional $f_x$ such that $f_x(y)=\langle x,y\rangle$, it is simply
$$ x^T=\begin{pmatrix}x_1&x_2&x_3\end{pmatrix}$$
That the transpose of a vector defined in this way exists in infinite dimensional Hilbert spaces is not obvious, and it's guaranteed by Riesz Representation Theorem. Basically, given a vector $x$, we define $x^T$ as the element of the dual such that $x^Ty=\langle x,y\rangle$, then this $^T$ map from the space to its dual is an isomorphism.
As a side note that may or may not be interesting:
in quantum mechanics, for example, this is particularly useful because we want $x$ (denoted $|x\rangle$) and $x^T$ (denoted $\langle x|$) to represent the same physical thing. Thanks to RRT it makes sense, given two vectors $|x\rangle$ and $|y\rangle$, to view their inner product as the product of $\langle x|$ with $|y\rangle$, or $\langle x|y\rangle$
A: You are reading the theorem wrong, I think, or are giving its claims the wrong weight, emphasizing the trivial direction. The important claim is that if you take a linear functional $\alpha\in X^*$, then there exists a vector $x$ so that the function values of the linear functional are the scalar products with $x$, $\alpha(y)=\langle x,y\rangle$, or $\alpha=f_x$.
Geometrically it can be interpreted that for the hyperplane $\{y:\alpha(y)=0\}$ there is one (and only one) normal direction. This is trivial in Euclidean geometry, but not as trivial in infinite dimensional Hilbert spaces.
A: I would answer this in two parts. This is in no way complete but it is what brought the theorem closer to me.
1) What does the theorem say?
First we need to know that for every vector space $V$ over a field $F$ there is a dual space called $V^*$ which consists of all linear maps $\varphi : V \to F $. This is a separate fact. 
What the theorem says now is every such $\varphi$ has a $v\in V$ that basically can be used in it's place if you want, meaning $\varphi(x) = <v,x>$. This has the implication, that there is a one to one relationship between all elements of $V^*$ and  $V$ making it possible get to the second part.
2) What are Riesz representers (as you call them)?
In a way it's legitimate to interpret them as elements of $V$, because of the isomorphic correspondence. 
They are something different though, which is the reason in Quantum mechanics if we have a complex Hilbert space $H$ and $v\in H$ we often write $v$ as $ |v> $ and the corresponding $\varphi \in V^*$ as $<v|$, marking that is "belongs to $v$" but it's not the same, really.  
If your interest in this comes from QM I can recommend the chapter in Ballentines Quantum Mechanics a Modern Development on it. 
A: As Lutz Lehmann already pointed it out, the important part of the theorem is: if you take a Hilbert space $H$ and a continuous linear functional on $H$, let's call it $f$, you have an unique vector $x$ in $H$ so that $f(y)=(y|x)$ for all $y$ in $H$.
This means that the continuous dual of a Hilbert space is basically the same as the space itself. A good example for this is $\mathbb{R}^n$, where it means that the row and the coulomn vectors are basically the same thing.
Another important consequence is that if I give you a Hilbert space, you can "easily" list all of the continuous linear functionals on the vector space. However, if I give you a Banach space, you can't really show me a non-zero, continuous linear functional (Of course, you can show that they exists with the axiom of choice).
