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Updated improved question:

Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define $$ \langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1). $$

(i) Prove that $\langle\cdot,\cdot\rangle$ defines an inner product.

Could we just do this $f(a)=0$ and $f'(a)=0$ then $f(x)$ is divisible by $(x-a)^2$ ?

If so how would we solve this?

Can someone please help me with this proof for part (i). It is frustrating me.

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  • $\begingroup$ Can you let us know where you are getting stuck? The definition of an inner product gives you a list of properties that a map $V \times V \to \mathbb{R}$ must satisfy. Think of this list like a check list, where you confirm that each property on the list is satisfied. $\endgroup$ – Tom Nov 5 '13 at 2:53
  • $\begingroup$ well, you have to show linearity in the first variable, symmetry and that it is positive definite. Which one is giving problems? $\endgroup$ – user67133 Nov 5 '13 at 2:53
  • $\begingroup$ @user01123581321345589144... ok I see. Maybe that is what I was forgetting. I did something like that but forgot to show it was positive definite. Thanks. $\endgroup$ – user99744 Nov 5 '13 at 3:00
  • $\begingroup$ @Tom I did post something like this before and showed the properties to each. I do not know if you saw it. It was very close but off a bit. I just wanted to see what other solutions there would be. I get it though. Thanks. $\endgroup$ – user99744 Nov 5 '13 at 3:02
  • $\begingroup$ I think this is the answer $$ \langle p,p\rangle = p(1)^2 + p'(1)^2 + p''(1)^2 + p'''(1)^2 \geq 0 $$ And if $\langle p,p\rangle = 0$, then note that $$ p(1) = p'(1) = p''(1) = p'''(1) = 0 $$ Now write $$ p(t) = at^3 + bt^2 + ct + d $$ and $a=b=c=d=0$ and $$ p = 0 $$ $\endgroup$ – user99744 Nov 5 '13 at 3:05
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The function you have defined is clearly bi-linear and symmetric, so the only thing one needs to check is positive-definiteness. For this, note that $$ \langle p,p\rangle = p(1)^2 + p'(1)^2 + p''(1)^2 + p'''(1)^2 \geq 0 $$ And if $\langle p,p\rangle = 0$, then note that $$ p(1) = p'(1) = p''(1) = p'''(1) = 0 $$ Now write $$ p(t) = at^3 + bt^2 + ct + d $$ and see that $a=b=c=d=0$ and conclude that $$ p = 0 $$

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  • $\begingroup$ Thank you. I understand the last part of it now. $\endgroup$ – user99744 Nov 6 '13 at 2:44
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Hint: write $p(x)=ax^3+bx^2+cx+d$. What are the derivatives of $p$ at $1$ with respect to $a,b,c,d$?

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In order to be an inner product, some conditions are necessary:

  • You need $\langle f + g, h\rangle = \langle f, h \rangle + \langle g, h \rangle$. Just write out each side and do some algebra.

  • You need $\langle \alpha f, h\rangle = \alpha \langle f, h\rangle$ for constants $\alpha$. Again just expand both sides and compare.

  • You need $\langle f, g \rangle = \langle g, f \rangle$. This is immediate by the same reasoning.

  • You need $\langle f, f \rangle = 0 \implies f = 0$. Suppose that $f(x) = ax^3 + bx^2 + cx + d$ is any polynomial of degree at most $3$. Then

\begin{align*} \langle f, f \rangle = (a + b + c + d)^2 + (3a + 2b + c)^2 + (6a + 2b)^2 + (6a)^2 \end{align*} The only way you can add a bunch of non-negative numbers and get zero is if each was already $0$. So we see that $6a = 0$, $6a + 2b = 0$, and so on. Can you show that $f = 0$?

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