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(i) If $f,g \in C^{0}(a,b)$, show that $$(f,g) = \int_a^b f(x)g(x)dx$$ is an inner product and $$\|f\|_{L^2(a,b)} = \int_a^b|f(x)|^2 dx$$ is a norm.

(ii) Determines the polynomials of degree $2$ orthogonal to $p_0(x) = 1$ and $p_1(x) = x$ in $L^2(-1,1)$.

(iii) Calculate an orthonormal basis of $\mathcal{P}_2(-1,1)$ from the direct product that generates the norm $L^2(-1,1)$ and of polynomials $\{1,x,x^2\}$.

Attempt.

(i) I think that the correct is $\sqrt{\int_a^b|f(x)|^2 dx}$.

(ii) We have that $$(p,p_0) = \int_{-1}^{1}p(x)dx = \int_{-1}^{1}(ax^2 + bx + c)dx = \left[a\frac{x^3}{3} + b\frac{x^2}{2} + cx\right]_{-1}^{1} = \frac{2a}{3} + 2c$$ and $$(p,p_1) = \int_{-1}^{1}p(x)xdx = \int_{-1}^{1}(ax^3 + bx^2 + cx)dx = \left[a\frac{x^4}{4} + b\frac{x^3}{3} + c\frac{x^2}{2}\right]_{-1}^{1} = \frac{2b}{3}.$$

Thus, we should have $b = 0$ and $a = -3c$. Hence the polynomials are $$\{-3cx^2 + c: c \in \mathbb{R}\}.$$

(iii) The item (ii) give us an orthogonal basis of $\mathcal{P}_2$, but if this item was not in question, I would use Gram-Schmidt. Take $u_1 = 1$ as the inicial vector. Then $$u_2 = x - \frac{(x,1)}{(1,1)}1 = x$$ $$u_3 = x^2 -\frac{(x^2,1)}{(1,1)}1 - \frac{(x^2,x)}{(x,x)}x = x^2 - \frac{1}{3}.$$

So, $\left\{\frac{1}{\|1\|},\frac{x}{\|x\|},\frac{-3x^2 + 1}{\|-3x^2 + 1\|}\right\}$ is an orthogonal basis.


I have another question about direct product.

The set $\{(1,1),(2,-1)\}$ is an orthonormal basis of $\mathbb{R}^2$. Which is the inner product?

Actually, that basis is not orthonormal, but I think that it is a typo or something like that. Anyway, I cannot see how the inner product depends on basis. Can someone help me?

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    $\begingroup$ What do you mean by direct product? $\endgroup$ – Rick Sanchez Oct 1 at 22:07
  • $\begingroup$ @RickSanchez I've corrected $\endgroup$ – Greg Oct 1 at 22:11
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Given a basis (spanning set that is linearly independent) $v_1,\ldots, v_n$ of $\mathbb R^n$, we can always define a unique inner product that makes them orthonormal. Since it's a basis, for any $x,y \in \mathbb R^n$, we can find unique constants such that $$x=\sum_1^n a_iv_i ,\,\, y=\sum_1^n b_iv_i.$$ Define $\langle x,y\rangle$ by $$\langle x,y\rangle = \sum_1^n a_ib_i.$$ You can check your basis is orthonormal with respect to this inner product.

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  • $\begingroup$ So, the question asks for a inner product that makes $\{(1,1),(2,-1)\}$ orthonormal? $\endgroup$ – Greg Oct 1 at 22:16
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    $\begingroup$ Yes, the point is you need to find the constants and then the above construction is the only answer. $\endgroup$ – Rick Sanchez Oct 1 at 22:19

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