Find the polynomial in a linear subspace that minimizes the distance from a given polynomial

Question

Consider the inner product space $\mathcal{P}[0,1]$ of all real polynomials on [0,1] with inner product $\langle f,g\rangle =\int_{0}^{1}f(t)g(t)dt$ and $V=\operatorname{span}\{t^2\}$. Let $h(t)\in V$ be such that $\|2t-1-h(t)\| \le \|2t-1-x(t)\|$ $\forall x(t) \in V $. Then $h(t)$ is

(a) $\frac{5}{6} t^{2}$

(b)$\frac{5}{3} t^{2}$

(c)$\frac{5}{12} t^{2}$

(d)$\frac{5}{24} t^{2}$

Let $h(t)=\alpha t^2$, $g(\alpha)=\|2t-1-h(t)\|$, Expanding the the expression, $g(\alpha)=\langle 2t-1-h(t), 2t-1-h(t) \rangle$

$= \int_{0}^{1}(2t-1-\alpha t^3)^2dt$. Then Extremises $g(\alpha)$. I got $h(t)=\frac{5}{3} t^{2}$.

Am I correct?

Can I do the problem in different method? Please suggest all the other possible ways.

Answer 1

You want the closest point projection of $2t-1$ onto the line $\{ \alpha t^2 : \alpha\in\mathbb{R} \}$. The orthogonal projection and the closest point projection are the same, which determines $\alpha$ by the equation $$ \langle 2t-1-\alpha t^2,t^2\rangle =0 \\ \alpha \langle t^2,t^2\rangle = \langle 2t-1,t^2\rangle \\ \alpha = \frac{\langle 2t-1,t^2\rangle}{\langle t^2,t^2\rangle} = \frac{\int_{0}^{1}(2t-1)t^2dt}{\int_{0}^{1}t^4dt}=\frac{\frac{2}{4}-\frac{1}{3}}{\frac{1}{5}}=\frac{5}{6}. $$ Looks like answer $(a)$ is correct.

Answer 2

You've probably made a slight calculation error but you're on the right track. You should find that $$ \int_0^1 (2t-1 - \alpha t^2)^2 dt = \frac{1}{15} \left( 3\alpha^2 - 5\alpha+5\right) $$ Minimising this should give you $\alpha = \frac{5}{6}$.

You can also easily find the answer if you take the projection of $f(t) =2t-1$ onto $V$. This projection is given by $$ Pf = \frac{\langle f,g \rangle}{\langle g,g \rangle} g $$ where $g(t) = t^2$. Notice that for any $\alpha \in \mathbb{R}$ we have $$ \| f-Pf \|^2 = \| (f-\alpha g) - (Pf- \alpha g)\|^2= \|f-\alpha g\|^2 - 2 \langle f - \alpha g, Pf - \alpha g \rangle + \|Pf- \alpha g \|^2 \\ = \| f-\alpha g \|^2 - \left( \frac{ \langle f, g \rangle}{\langle g , g \rangle} - \alpha \right)^2 \langle g,g \rangle. $$ So clearly $\| f-Pf \| \leq \| f - \alpha g \|$ for all $\alpha \in \mathbb{R}$. Then all that is left to do is calculate $$ \frac{\langle f, g \rangle}{\langle g, g \rangle} $$ which is indeed $\frac{5}{6}$ as we found earlier.