7
$\begingroup$

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

  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.

$\endgroup$
  • 1
    $\begingroup$ If 2-5 are true, can you say $s(x+y,z)=s(x(1+y/x),z)=(1+y/x)s(x,z)=s(x,z)+(y/x)s(x,z)=s(x,z)+s(y,z)$? $\endgroup$ – Empy2 Mar 14 '16 at 23:22
  • 1
    $\begingroup$ @Michael: Only if $1+y/x\in K$. $\endgroup$ – joriki Mar 14 '16 at 23:22
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.