Im currently trying to find a nonzero function $h(x)$ in span $\{1,x,x^2\}$ such that $\langle h,1\rangle = 0$ and $\langle h,x\rangle = 0$ where the inner product is defined by $\langle f,g\rangle = \int_0^1 f(x)g(x)~dx$ but I'm having trouble coming up with one.
Does such an example exist?
My work: i put coefficients $a,b,c$ in front of the spanned terms and evaluated the two conditions. I'm left with $a + b/2 + c/3 = 0$ and $a/2 + b/3 + c/4 = 0$. I don't know where to go from here.
Suppose $h=k_1+k_2x+k_3x^2$, then you have$$\begin{align*}\int_0^1(k_1+k_2x+k_3x^2)\cdot1~dx=0&\iff k_1+k_2/2+k_3/3=0\\\int_0^1(k_1+k_2x+k_3x^2)\cdot x~dx=0&\iff k_1/2+k_2/3+k_3/4=0\end{align*}$$We can solve for $k_i$ as follows:$$k_1=-k_2/2-k_3/3=-2k_2/3-k_3/2\implies k_2+k_3=0$$Thus we get $k_1=-k_2/6,k_3=-k_2$. We may select any value for $k_2$, say $k_2=1,k_3=-1,k_1=-1/6$. In fact, the general quadratic that satisfies our requirement is a constant multiple of $(-1/6+x-x^2)$ as$$k_1+k_2x+k_3x^2=k_2(-1/6+x-x^2),k_2\in\Bbb R$$