Improved part to this 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.

Here is what I had:

Consider $P_t$, the space of polynomial of degree polynomial at most $t$ ( with coefficients from R)Let $x_0<x_1<x_2<…<x_3$ be any $t+1$ real numbers. For $p$ and q belong to $P_t$ ,we define $<p,q>$ to be $p(1)q(1)+p(1)'q(1)'+p(1)''q(1)''+p(1)'''q(1)'''$ is an inner product on $P_t$. Checking that $<p,P> = 0 $->$ p = 0$ then $1$ is the root of $p$, and as a polynomial of degree at most $t$ has at most $t$ roots unless it's the zero polynomial, p must be the zero polynomial. Different choice of $1$ gives rise to different inner products.

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

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


$$ \langle p,p\rangle = p(1)^2 + p'(1)^2 + p''(1)^2 + p'''(1)^2 \geq 0 $$ 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 $$ p = 0 $$


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.