Functions satisfying 4 out of 5 inner product properties

Question

Let us consider function $s:K^m \times K^m \mapsto K$ (here $K = \mathbb{R}$ or $K = \mathbb{C}$). If $\forall x, y, z \in K^m, \forall \lambda \in K$

1. $s(x + y, z) = s(x, z) + s(y, z)$
2. $s(\lambda x, y) = \lambda s(x, y)$
3. $s(y, x) = \overline{s(x, y)}$
4. $s(x, x) \geq 0$
5. $s(x, x) = 0 \implies x = 0$

then $s$ is called inner product.

Problem. For each $n = 1, 2, 3, 4, 5$ find a function $s$ that doesn't satisfy the $n$-th property and satisfies the remaining four.

First consider $K = \mathbb{R}$. I found the following:

$n = 3, s(x, y) = xy^3$

$n = 4, s(x, y) = -xy$

$n = 5, s(x, y) \equiv 0$

How can I approach $n = 1, 2$? Perhaps I need to choose $K = \mathbb{C}$ for those?

---

Edit: I changed the domain of $s$ from $\mathbb{R} \times \mathbb{R}$ to $K^m \times K^m$ because

1. if $\lambda \in \mathbb{C}$ then $\mathbb{R}$ is not closed w.r.t. scalar multiplication and

2. if $s: K \times K \mapsto K$ and 2-5 hold then 1 must hold.

Answer 1

Cases $n = 3, 4, 5$ have been shown in the OP.

---

Case $n = 1$.

I will show that there are no such functions from $K \times K$ and provide an example for $\mathbb{R}^2 \times \mathbb{R}^2$.

Let $m = 1$. Take property 2 and choose $x = 1$ so $s(\lambda, y) = \lambda s(1, y)$. Denote $s(1, y) = f(y)$ so $s(x, y) = x f(y) \ \forall x, y$. By symmetry property, $s(x, y) = \overline{y f(x)}$. Hence

$$\frac{f(y)}{\overline{y}} = \frac{\overline{f(x)}}{x} = c = \text{const}$$

because $x, y$ can be any elements of $K$.

This immediately gives $s(x, y) = c x \overline{y}$. It is easy to see that 3-5 hold, as well, if $c \in \mathbb{R}$ and $c > 0$. Inserting in 1, we see that it holds, as well.

So there are no such functions for $m = 1$.

Let $m = 2$, $K = \mathbb{R}$ and denote $x = (x_1, x_2) \in \mathbb{R}^2$. Let

$$s(x, y) = \sqrt[3]{(x_1 y_1)^3 + (x_2 y_2)^3}.$$

Obviously, properties 2-5 hold but 1 (additivity) doesn't.

---

Case $n = 2$.

If $K = \mathbb{C}$ then we can choose

$$s(x,y) = \overline{x}y$$

Property 1 holds: $\overline{(x + y)}z = \overline{x}z + \overline{y}z$

Property 2 doesn't hold if $\text{Im} \lambda \neq 0$: $\overline{\lambda x}y \neq \lambda \overline{x}y$

Property 3 holds: $\overline{\overline{x}y} = x\overline{y} = \overline{y}x$

Properties 4, 5 hold: $\overline{x}x = |x| \geq 0$ and $|x| = 0 \implies x = 0$

Note: for $K = \mathbb{R}$ there are functions that satisfy $f(x + y) = f(x) + f(y)$ but aren't linear (see here). If $h(x)$ is such a function, $s(x,y) = h(x)h(y)$ would't satisfy property 1 and would satisfy 2-4. I am unsure how to make it satisfy property 5.