Is this an inner product?

Question

This is a question from Apostol's Vol. 2 book on Calculus.

Does the following violate the definition on an inner product. The answer is it does. Please let me know which rule it violates and point out where the error is in my discussion. Thanks a lot.

Definition: $V$ is the linear space of all polynomials, and $f,g,h \in V$

$$ (f,g) = f(1)g(1) $$

1. Property: $(x,x) > 0 \text{ if x}\neq\text{O}$ $$ (f,f) = f(1)^2 > 0 \text{ if } f(1)>0 $$ Satisfied.
2. Property: $(x, y+z) = (x,y) + (x,z)$ $$ (f,g+h) = f(1)[g(1)+h(1)] = f(1)g(1) + f(1)h(1) = (f,g) + (f,h)$$ Satisfied.
3. Propert: $c(f,g) = (cf, g)$ $$ c(f,g) = cf(1)g(1) = (cf(1))g(1) = (cf,g) $$ Satisfied.
4. Property: $(f,g) = (g,f)$ $$ (f,g) = f(1)g(1) = g(1)f(1) = (g,f) $$ Satisfied.

All 4 properties of an inner product is satisfied, yet Apostol says it isn't.

Answer 1

The symmetry and the linearity are obviously satisfied, as you showed correctly. However, try to look at property 1 for the polynomial $f(x) = x - 1$. This polynomial is zero in $x=1$, yet it is definitely not the zero polynomial.

So as you wrote yourself, $(f,f) > 0$ only if $f(1) \neq 0$, but certainly not every polynomial satisfies this condition!

This is a typical problem for norms (and also scalar products): they have to somehow use "all the information". It is not enough to just look at the value in one single point!

Hope that helps!

Answer 2

Property (1) is not satisfied by $f(x) = x - 1$;

$f(1) = 1 - 1 = 0 \Longrightarrow (f(x), f(x)) = (f(1))^2 = 0, \tag{1}$

but $f(x) \ne 0$.