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Let $\psi: l^2 \to \mathbb{R}$, I can't understand what this stands for

$$<\psi, e_j>=\frac{(-1)^j}{j!}\in\mathbb{R}$$

$e_j$ are zero vectors with a $1$ in the $j$-th position.

What informations it gives me on $\psi$?

If $\psi$ was a sequence I would say that the product is the standard inner product of $l^2$ elements, so the result is the $j$-th element of the sequence $\psi$.

Otherwise $\psi$ is a function, what does it mean?

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$<\psi,x>$ is just another way of writing $\psi(x)$.Therefore the above expression gives you the value of the function at all of the $e_j's$. Moreover, if $e_j's$ are the basis elements and $\psi$ is a linear function then it completely defines $\psi$ on the vector space.

Hope it is helpful

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  • $\begingroup$ So $e_j$ are a basis of $l^p$? $\endgroup$ – james watt Nov 14 '18 at 10:53
  • $\begingroup$ What do you denote by $l^p?$ $\endgroup$ – Martund Nov 14 '18 at 10:55
  • $\begingroup$ The space of sequences with finite $l^p$ norm. $\endgroup$ – james watt Nov 14 '18 at 10:56
  • $\begingroup$ It's a Hilbert- but not a Schauder-basis (in the sense of Linear Algebra). $\endgroup$ – Michael Hoppe Nov 14 '18 at 11:17
  • $\begingroup$ @MichaelHoppe on wiki it is written "The spaces $c_0$ and $ℓ^p$ (for 1 ≤ p < ∞) have a canonical unconditional Schauder basis $\{e_i | i = 1, 2,…\}$, where $e_i$ is the sequence which is zero but for a $1$ in the i-th entry." $\endgroup$ – sound wave Nov 14 '18 at 11:20

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