Inner product problem

**arbolis** · June 5th 2008, 05:11 PM

Deduce in each cases, which of the following applications $\langle \cdot,\cdot \rangle$ : $S$ x $S\rightarrow$ $\mathbb{R}$ are inner products.
a) $\langle f,g\rangle=f(0)+f(1)+2g(0)$ , with $S=$ {Real polynomials with grade $\geq 1$ }
b) $\langle f,g\rangle=f(0)g(0)+f(1)g(1)$ with $S=$ {Real polynomials with grade $\geq 2$ }.
I've tried to start, but I'm not sure how to.
I know that in order to be an inner product, the application must agree with 4 condition, for example one of them is $\langle f,f\rangle$ must be strictly positive. I just don't understand enough the problem and I'm very new to all this (inner product, basis, etc.). Can you please help me?

**Opalg** · June 6th 2008, 12:46 AM

Originally Posted by arbolis

Deduce in each cases, which of the following applications $\langle \cdot,\cdot \rangle$ : $S$ x $S\rightarrow$ $\mathbb{R}$ are inner products.
a) $\langle f,g\rangle=f(0)+f(1)+2g(0)$ , with $S=$ {Real polynomials with grade $\geq 1$ }
b) $\langle f,g\rangle=f(0)g(0)+f(1)g(1)$ with $S=$ {Real polynomials with grade $\geq 2$ }.

Two preliminary comments. First, the usual English word is degree of a polynomial (not grade). Second, the set of polynomials with degree $\geq1$ , or $\geq2$ , is not a vector space. I suspect that you really mean $\leq1$ and $\leq2$ .

Originally Posted by arbolis

I've tried to start, but I'm not sure how to.
I know that in order to be an inner product, the application must agree with 4 condition, for example one of them is $\langle f,f\rangle$ must be strictly positive.

That is exactly the correct strategy. You have to look at the four axioms for an inner product, and decide in each case whether that axiom is satisfied. For example, look at the axiom $\langle f,f\rangle>0$ (for all f≠0). For problem a), this is saying that 3f(0)+f(1)>0 for all polynomials of degree 1. For problem b), it says that $f(0)^2+f(1)^2>0$ for all polynomials of degree 2. In each case, you have to decide whether or not this is true.

**arbolis** · June 6th 2008, 06:32 AM

I suspect that you really mean

and

.

Yes, sorry about that.

look at the axiom

(for all f≠0). For problem a), this is saying that 3f(0)+f(1)>0 for all polynomials of degree 1.

How do you know this? Did you do the inner product $\langle f,f\rangle$ ? I'm not able to do it since f is not given. What does $S$ x $S\rightarrow\mathbb{R}$ mean? Does that mean in the first example the inner product is a product of 2 polynomials of degree $\leq 1$ and that the result will be a real number? If yes, then I can rewrite $\langle f,f\rangle$ as $\langle ax+b,ax+b\rangle$ but from here I'm not able to do the inner product. Usually I know how to do the inner product when I have one of the form $\langle (a,b),(c,d)\rangle$ , so I need a coma in each term of the inner product. Can you help me a bit more?

**Opalg** · June 6th 2008, 06:44 AM

Originally Posted by Opalg

look at the axiom $\langle f,f\rangle>0$ (for all f≠0). For problem a), this is saying that 3f(0)+f(1)>0 for all polynomials of degree 1.

Originally Posted by arbolis

How do you know this? Did you do the inner product $\langle f,f\rangle$ ? I'm not able to do it since f is not given. What does $S$ x $S\rightarrow\mathbb{R}$ mean?

What this means is that the inner product is a mapping that takes two elements f, g of the space S and associates a real number $\langle f,g\rangle$ with that pair. For problem a), you are told that the formula for the inner product is $\langle f,g\rangle = f(0)+f(1)+2g(0).$ If you put g=f in that formula then you get $\langle f,f\rangle = 3f(0)+f(1).$

**arbolis** · June 6th 2008, 07:17 AM

For problem a), you are told that the formula for the inner product is

If you put g=f in that formula then you get

Oh...! Sometimes I can't see the obviousness. Thanks a lot for this, this was worthful. Thus now I must show whether or not $3f(0)+f(1)>0$ , knowing that f is a polynomial of degree $\leq 1$ . If so, then it's false. As a counter example I can take $f(x)=-10x-10^9$ . So the a) wouldn't be an inner product.
For the b), we have $f(0)^2+f(1)^2>0$ as you said, which is always true if f is different from 0.
Now I must check if it also agrees with the 3 other conditions.
There is the condition $\langle f,g\rangle=\langle g,f\rangle$ . It also works.
Another one would be $\langle af,g\rangle=a\langle f,g\rangle$ where a is a constant. It also works!!!
From wikipedia Inner product space - Wikipedia, the free encyclopedia, the fourth condition would be $\langle f,f\rangle=0 \Rightarrow f=0$ . Which works. So b) is an inner product.
Can you tell me if I used the good conditions?

**Opalg** · June 6th 2008, 07:31 AM

Originally Posted by arbolis

Oh...! Sometimes I can't see the obviousness. Thanks a lot for this, this was worthful. Thus now I must show whether or not $3f(0)+f(1)>0$ , knowing that f is a polynomial of degree $\leq 1$ . If so, then it's false. As a counter example I can take $f(x)=-10x-10^9$ . So the a) wouldn't be an inner product.
For the b), we have $f(0)^2+f(1)^2>0$ as you said, which is always true if f is different from 0.
Now I must check if it also agrees with the 3 other conditions.
There is the condition $\langle f,g\rangle=\langle g,f\rangle$ . It also works.
Another one would be $\langle af,g\rangle=a\langle f,g\rangle$ where a is a constant. It also works!!!
From wikipedia Inner product space - Wikipedia, the free encyclopedia, the fourth condition would be $\langle f,f\rangle=0 \Rightarrow f=0$ . Which works. So b) is an inner product.
Can you tell me if I used the good conditions?

That's very nearly correct now. The only thing that still looks wrong is that a nonzero quadratic polynomial can vanish at x=0 and x=1. If, say, $f(x)=x^2-x$ , then $f(0)^2+f(1)^2=0$ although f is not the zero polynomial. So my conclusion is that b) also fails to be an inner product.

**arbolis** · June 6th 2008, 07:41 AM

The only thing that still looks wrong is that a nonzero quadratic polynomial can vanish at x=0 and x=1.

You're right, I should have took more care. So the conclusion follows are you said. Thanks for your help.

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