I'm having some trouble proving the following theorem:
Let $($$X$,$\langle\cdot | \cdot\rangle$$)$ be an inner product space and $f: X \to \mathbb{R}$ a linear mapping. Prove that there exists a unique $\boldsymbol{\tilde{x}} \in X$ such that $f(\boldsymbol{x})=\langle\boldsymbol{x} | \boldsymbol{\tilde{x}}\rangle$ $\forall \boldsymbol{x} \in X$.
Some tips would be great!
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I can answer for the finite-dimensional case. I'm not that experienced with infinite dimensional inner product spaces. I'm hoping the argument won't change much for the infinite dimensional case.
For the finite dimensional case with $C^n$ as the underlying field, let $e_1,\ldots ,e_n$ be an orthonormal basis for $C^n$.
Thus, we have $x = \langle x, e_1 \rangle e_1 +\ldots + \langle x, e_n \rangle e_n $
Then from the linearity of inner products and $f$, $$ \begin{align} f(x) &= f\left(\langle x, e_1 \rangle e_1 +\ldots + \langle x, e_n \rangle e_n \right) \\ &= \langle x, e_1 \rangle f(e_1) +\ldots + \langle x, e_1 \rangle f(e_2) \\ &= \langle x, \overline {f(e_1)} e_1 \rangle + \ldots+\langle x, \overline {f(e_n)} e_n \rangle \\ &= \langle x, \overline {f(e_1)} e_1 +\ldots+\overline {f(e_n)} e_n\rangle \end{align} $$
Thus, $f(x) = \langle x, u \rangle$, where $u = \overline {f(e_1)} e_1 +\ldots+\overline {f(e_n)} e_n$ .
For general inner product spaces, this statement is wrong. Consider for exmaple, the space $X = c_{00}$, that is $$ X = \{(x_n) \in \mathbf R^{\mathbf N} \mid \exists N \, \forall n \ge N: x_n = 0\} $$ the finitely supported sequences, with the inner product $$ \def\<#1|#2>{\left<\left.#1\right|#2\right>} \<x|y> := \sum_n x_n y_n $$ Define $f \colon X \to \mathbf R$ by $$ f(x) := \sum_n x_n $$ Then $f$ is obviously linear, but is not of the desired form: Let $y \in X$ be any point, we will show $f \ne \<\cdot|y>$. Choose $N$ such that $y_n = 0$ for $n \ge N$ and let $x = e_N$, that is $x_N = 1$ and $x_n = 0$ for $n \ge N$. Then $$ f(x) = 1 \ne 0 = \sum_n x_n y_n = \<x|y>. $$ It is true for complete inner product spaces, where it can be deduced from the projection theorem.