How we can find the orthogonal complement $h \in W^{⊥}$

Let $$V$$ be the vector space of polynomials in the variable $$t$$ of degree at most $$2$$ over \mathbb{R}. An inner product on $$V$$ is defined by $$=\int_{0}^{1} f(t)g(t)dt$$ for $${f,g} \in V$$. Let $$W=span\{1-t^{2}, 1+t^{2}\}$$ and $${W^{⊥}}$$ be the orthogonal complement of $$W$$ in $$V$$. Which of the following conditions is satisfied for all $$h \in{W^{⊥}}$$ ?

1. $$h$$ is an even function.
2. $$h$$ is an odd function.
3. $$h(t)=0$$ has a real solution.
4. $$h(0)=0$$.

Orthogonal complement $$W^{⊥}= \{x\in V | =0 \, y\in W\}$$. Please give some hint.

• The downvote (wasn't me) is probably because you didn't tell us what you tried or found. You should add that if possible Nov 12 '19 at 15:02
• I know the orthogonal complement definition .But how I approach to this..... Nov 12 '19 at 15:09
• Try taking an arbitrary $h\in V$ and taking the inner product with the basis vectors of $W$. What does that tell you? Nov 12 '19 at 15:12
• Then for finding $h(t)$ these should be hold $\int_{0}^{1} (1-t^{2})h(t)dt=0, \int_{0}^{1} (1+t^{2})h(t)dt=0$ Nov 12 '19 at 15:19
• If you get stuck again, since no one answered yet you can probably expand your question with what you've found and where you're stuck. If you solve the problem, you can make an answer yourself. Nov 12 '19 at 15:24

You should consider an $$h\in V$$, and take the inner product with the basis vectors of $$W$$. The inner products $$\int_0^1(1\pm t^2)(at^2+bt+c)\, dt$$ should evaluate to zero. They evaluate to $$\frac{8a}{15}+\frac{3b}{4}+\frac{3c}{2} = 0$$ and $$\frac{2a}{15}+\frac{b}{4}+\frac{c}{2} = 0$$ for the plus and minus respectively. Subtracting three times the second one from the first yields $$2a/15=0$$. Subtracting four times the second one from the first yields $$-b/4-c/2=0$$. We get $$W^\perp=\mathrm{span}\{-2t+1\}$$. That means only the third condition is always satisfied.