I know Riesz Representation Theorem in Hilbert spaces.
If $T$ is a bounded linear functional on a Hilbert space $H$ then there exists some $g \in H$ such that for every $f \in H$ we have
$$T(f) =\langle f,g\rangle.$$
Moreover, $\|T\| = \|g\|$ (here $\|T\|$ denotes the operator norm of $T$, while $\|g\|$ is the Hilbert space norm of $g$).
I need couple of examples on different Hilbert spaces.
We know $g$ exists and it is unique. I am looking for example and finding $g$, for example, if we consider $\Bbb{R}^2$ as our space.
Also I am wondering if I can find all bounded linear operators using Riesz Representation Theorem. To be more specific, if I consider a $g$ using the dot product, I can generate a bounded operator.
Best Answer
If you're working in $\Bbb{R}^n$, then you have the standard basis $e_1, \ldots, e_n$. Then given a linear functional $T$, \begin{align*} T(x_1, \ldots, x_n) &= T(x_1 e_1 + x_2 e_2 + \ldots + x_n e_n) \\ &= x_1 T(e_1) + x_2 T(e_2) + \ldots + x_n T(e_n) \\ &= (x_1, \ldots, x_n) \cdot (T(e_1), \ldots, T(e_n)). \end{align*} Thus, we have a vector in $\Bbb{R}^n$ which represents $T$. Is it the only one? Let's suppose that $(u_1, \ldots, u_n) \in \Bbb{R}^n$ also represents $T$. That is, for all $(x_1, \ldots, x_n) \in \Bbb{R}^n$, $$T(x_1, \ldots, x_n) = (x_1, \ldots, x_n) \cdot (u_1, \ldots, u_n).$$ Plugging in $e_i$, we get, $$T(e_i) = 0u_1 + 0u_2 + \ldots + 0u_{i-1} + 1 u_i + 0u_{i+1} + \ldots + 0u_n = u_i.$$ Therefore $u_i = T(e_i)$ for all $i$, and so $$(u_1, \ldots, u_n) = (T(e_1), \ldots, T(e_n)),$$ i.e. the representative is unique.