Why does $\langle x | f \rangle = f(x) $?

Question

i am trying to understand why does $\langle x | f \rangle = f(x) $.

My line of thought so far:

we will write $f_n(x)$ as the discrete approximation of $f(x)$, where $f_n(x)$ is given by $| f_n \rangle = \begin{pmatrix} f_n(x_1) \\ \vdots \\ f_n(x_n) \end{pmatrix} $, $| x_i \rangle = \begin{pmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{pmatrix} $, $|f_n\rangle=\sum_{i=1}^{n} f_n(x_i) \ |x_i\rangle$, and $f_n(x_i)=\langle x_i|f_n\rangle$. So we can now say that if we take an infinite number of points, we get $|f\rangle $ as an infinite series, and $|x\rangle $ is also an infinite column vector which when applied to $f$ will give us only its value at $x$ (i.e. $f(x)$), as did $|x_i\rangle$ with $|f_n\rangle$.

Is this line of thought correct? could you make it more precise and formal?

Answer 1

In Shankar's book, the point is to determine the components of $|\psi\rangle$ in the position basis. If we label the components as $\psi(x)$, then we want

$$|\psi\rangle=\int\psi(x')|x'\rangle\ \text dx'$$

Since $\langle x|x'\rangle=\delta(x'-x)$, we get

$$\langle x|\psi\rangle=\int\psi(x')\langle x|x'\rangle\ \text dx'=\psi(x)$$

But I think this is kind of circular, or at least there are different "points of entry". If we say $\langle x|\psi\rangle$ will always pick out the "x component" of $|\psi\rangle$, and this will be a function of $x$, so we label it as $\psi(x)$ for clarity. In other words.... if we set up the notation $\langle x|\psi\rangle$ to be the "x component" of $|\psi\rangle$ understood to be this linear combination, then it kind of just falls out if we assume a orthonormal position basis.