# Does the inner product define a function?

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?

$$<\psi,x>$$ 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.

• So $e_j$ are a basis of $l^p$? – james watt Nov 14 '18 at 10:53
• What do you denote by $l^p?$ – Martund Nov 14 '18 at 10:55
• The space of sequences with finite $l^p$ norm. – james watt Nov 14 '18 at 10:56
• It's a Hilbert- but not a Schauder-basis (in the sense of Linear Algebra). – Michael Hoppe Nov 14 '18 at 11:17
• @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." – sound wave Nov 14 '18 at 11:20