How to find $W^{\perp}$ in the following polynomial inner product space?

Question

Consider $P_3(\Bbb{R})$ with inner product $\langle p(x),q(x)\rangle=\int^1_{-1} p(x)q(x)dx$ and let $W=\{ p(x)\in P_3(\Bbb{R})|p(0)=p'(0)=p''(0)=0\}$. How to find $W^{\perp}$?

Let's set $p(x)=a+bx+cx^2+dx^3$. Since $p(0)=p'(0)=p''(0)=0$, $p(x)=dx^3$. To find $q(x)$, we need to find the $q(x)$ that satisfies $\langle p(x),q(x)\rangle=\int^1_{-1} p(x)q(x)dx=0$. However, the problem is if we let $q(x)=e+fx+gx^2+hx^3$ and multiply it by $p(x)$, then the subsequent $\langle p(x),q(x)\rangle=\int^1_{-1} p(x)q(x)dx=0$ would be unsolvable.

Could someone suggest a valid solution?

Answer 1

Since $$ \langle dx^3,q\rangle=\int^1_{-1}d(fx^4+hx^6)\,dx=d\left(\frac{2f}5+\frac{2h}7\right)=0 $$ for all $d$, you get $$ \frac{2f}5+\frac{2h}7=0.\tag1 $$ In other words, $W^{\bot}$ is the set of all polynomials $q$ satisfying (1).

Answer 2

a basis for the space $W=\{ p(x)\in P_3(\Bbb{R})|p(0)=p'(0)=p''(0)=0\}$ is $\{x^3\}.$ suppose $a + bx + cx^2 + dx^3$ is orthogonal to $x^3,$ then
$$0=a \int_{-1}^1 x^3dx+b\int_{-1}^1 x^4 \, dx+ c\int_{-1}^1x^5 \, dx + d \int_{-1}^1x^6\, dx=\frac {2b}5+\frac{2d}7$$ that means we can have $a, d$ arbitray and $b = -7, c = 5.$ that is a basis for the $W^\perp $ is $\{ 1, x^3, -7x + 5x^3\}$ of dimension $3.$

Answer 3

You stopped too soon; or, maybe you forgot where you were going. Since you're putting only one linear condition on the original function, you should expect a single linear condition on the coeffs. Just do the integration, and set the result to zero.