# Find a function that is orthonormal to both $g(x)$ and $h(x)$?

Find a function of the form $\ f(x) = ax^2+bx+c$ that is orthonormal to both $\ g(x) = 1$ and $\ h(x) = 1+x$ in $\ L_2(-1,1)$?

• Write the orthogonality condition, solve for coefficients. Commented Nov 15, 2017 at 5:51
• Hi. In order to get some responses, you should show what you have tried. Let others know what you have difficulties with. Otherwise you might get downvoted or have your question closed. Commented Nov 15, 2017 at 6:04
• You could also perform the Gram-Schmidt procedure on $1, 1 + x, x^2$. The last vector in the resulting list will be orthogonal to $1$ and $1 + x$, while still belonging to the span of $1, 1+x, x^2$, which will still be in the form $ax^2 + bx + c$. Commented Nov 15, 2017 at 6:40
• At least try something. Commented Nov 15, 2017 at 6:46
• @Andrei I tried assuming arbitrary coefficients and applying the orthogonality. Commented Nov 16, 2017 at 6:24

Conditions for orthogonality $$\int_{-1}^1\,f(x)g(x)\,dx=0;\;\int_{-1}^1\,f(x)h(x)\,dx=0$$ Condition for normality $$\int_{-1}^1\,f^2(x)\,dx=1$$

$$\int_{-1}^1\,(ax^2+bx+c)\,dx=\left[\frac{ax^3}{3}+\frac{bx^2}{2}+cx\right]_{-1}^1=\frac{a}{3}+\frac{b}{2}+c-\left(-\frac{a}{3}+\frac{b}{2}-c\right)=\frac{2a}{3}+2c$$

$$\int_{-1}^1\,(ax^2+bx+c)(x+1)\,dx= \frac{2}{3} (a + b + 3 c)$$

$$\int_{-1}^1\,(ax^2+bx+c)^2\,dx= \frac{2a^2}{5}+ \frac{4 a c}{3} + \frac{2 b^2}{3} + 2 c^2$$

We get the system

$\left\{ \begin{array}{l} \frac{2a}{3}+2c=0\\ \frac{2}{3} (a + b + 3 c)=0\\ \frac{2a^2}{5}+ \frac{4 a c}{3} + \frac{2 b^2}{3} + 2 c^2=1 \end{array} \right.$

simplified

$\left\{ \begin{array}{l} a+3c=0\\ a + b + 3 c=0\\ 6a^2+20ac+10b^2+30c^2=15 \end{array} \right.$

$\left\{ \begin{array}{l} a=-3c\\ b=0\\ 6a^2+20ac+30c^2=15 \end{array} \right.$

$\left\{ \begin{array}{l} a=-3c\\ b=0\\ 54c^2-60c^2+30c^2=15\to 24c^2=15 \to c^2=\frac{5}{8} \end{array} \right.$

two solutions $\left(-3\sqrt{\frac{5}{8}},0,\sqrt{\frac{5}{8}}\right);\;\left(3\sqrt{\frac{5}{8}},0,-\sqrt{\frac{5}{8}}\right)$

Hope this helps