(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 $\{1,x,x^2\}$.


(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?

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ So, the question asks for a inner product that makes $\{(1,1),(2,-1)\}$ orthonormal? $\endgroup$ – Greg Oct 1 at 22:16
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.