Inner Product of Real Polynomials

Question

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.

Answer 1

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 $$ p = 0 $$

Answer 2

Hint: write $p(x)=ax^3+bx^2+cx+d$. What are the derivatives of $p$ at $1$ with respect to $a,b,c,d$?

Answer 3

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 $f = 0$?