I have a task in the functional analysis exam that I could not do.

Problem: In Hilbert space $L_2 [0,1]$ (with Lebesgue measure) let:

$$V = \left\{ f \in L_2 [0,1]: \int_{0}^{1} t \cdot \overline{f(t)} = 0 = \int_{0}^{1} (5t^3 + 3t - 2) \cdot \overline f(t) \right\}$$

Find orthogonal basis for $V^{\perp}$ and find orthogonal projection $9t^2 + 2$ on $V$.

My attempt

From definition of $V$ I can write that $V^{\perp} = \hbox{span} \{t, 5t^3 + 3t - 2 \}$. Gram-Shmidt algorithm gives me orthogonal basis contains functions $f(t) = t$ and $f(t) = 5t^3 - 2$.

Now I have to write $9t^2 + 2 = a \cdot t + b \cdot (5t^3 - 2) + \hbox{rest}$. And the rest will by my answer. But I have no idea how can I evaluate it.

  • 1
    $\begingroup$ If you have an orthoNORMAL basis $e,f$ of $V^\perp$ the orthogonal projection to $V^\perp$ of $g\in L_2$ is P(g)=\langle g,e\rangle e + \langle g,f \rangle f$. $\endgroup$ – Jochen Sep 4 '18 at 10:07

Divide the basis functions $t$ and $5t^3-2$ with their norms to arrive at an orthonormal basis $\{e_1, e_2\}$ for $V^\perp$ where

$$e_1(t) = \frac1{\sqrt2}t, \qquad e_2(t) = \sqrt{\frac7{18}}(5t^3-2)$$

Now the orthogonal projection $P$ onto $V^\perp$ is given by

$$Pf = \langle f, e_1\rangle e_1 + \langle f, e_2\rangle e_2$$

and the orthogonal projection on $V$ is given by $I - P$.

The desired projection of $f(t) = 9t^2+2$ onto $V$ is then given by \begin{align} f - Pf &= f- \langle f, e_1\rangle e_1 - \langle f, e_2\rangle e_2\\ &= 9t^2+2 - \underbrace{\left(\int_0^1(9u^2+2)\frac1{\sqrt{2}}u\,du\right)}_{=\frac{13}{4\sqrt{2}}}\frac1{\sqrt{2}}t - \underbrace{\left(\int_0^1(9u^2+2)\sqrt{\frac7{18}}(5u^3-2)\,du\right)}_{=0}\sqrt{\frac7{18}}(5t^3-2) \\&= 9t^2 - \frac{13}8t + 2 \end{align}


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