# Norm Minimization Examples 1

Recall from the Norm Minimization page that if $V$ is an inner product space and $U$ is a finite-dimensional vector space of $V$ where $V = U \oplus U^{\perp}$, and if we let $v \in V$ then for every vector $u \in U$ we have that:

(1)

Furthermore, evaluating holds if and only if we take $u = P_U(v)$.

We will now identify how to use the theorem above in order to solve various minimization problems.

Step 1 | We must first identify a vector space $V$ and define an inner product space on $V$. |
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Step 2 | We will want to also identity the subspace $U$ of $V$ for which our solution lies in. We must note that $U$ must be a finite-dimensional vector space. |

Step 3 | We must then find a basis of $U$. Using the Gram-Schmidt procedure, we can then convert this basis to an orthonormal basis, $\{ e_1, e_2, …, e_n \}$ of $U$ with respect to the inner product defined on $V$. |

Step 4 | In applying Theorem 1 from above, we will have that the normed difference, $\| v – P_U(v) \|$ will be less than or equal to the normed difference of $\| v – u \|$ for every vector $v \in V$, that is, $\| v – P_U(v) \| ≤ \| v – u \|$. We choose $u = P_U(v)$ to minimize the difference between every vector in $V$. |

Step 5 | The vector $P_U(v) = |

Let’s look at an example.

## Example 1

**Find the polynomial $p(x) \in \wp_3 (\mathbb{R})$ that minimizes $\int_0^1 \mid 2 + 3x – p(x) \mid^2 \: dx$ such that $p(x) \in \wp_3 ( \mathbb{R})$ and $p(0) = 0$ and $p'(0) = 0$.**

We first identity that the larger vector space we’re looking at is $V = \wp_3 ( \mathbb{R} )$. Let $q(x) = 2 + 3x$. We define an inner product of $V$ by:

(2)

= \int_0^1 p(x) q(x) \: dx \end{align}

We’re looking for a polynomial in $\wp_3 (\mathbb{R})$ such that $p(0) = 0$ and $p'(0) = 0$. The set of polynomials of degree less than or equal to $3$ that satisfy these conditions are a subspace $U$ of $V$ defined by:

(3)

We will now find a basis of $U$. Let $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$. Note that $p(0) = a_0$, but we must have that $p(0) = 0$. Therefore $a_0 = 0$. Differentiate $p$ to get that $p'(x) = a_1 + 2a_2x^2 + 3a_3x^2$. Note that $p'(0) = a_1$, but we must have that $p'(0) = 0$, which implies that $a_1 = 0$.

Therefore, $p(x) = x_2a^2 + x_3a^3$, and so:

(4)

We can clearly see that $U = \mathrm{span} (x^2, x^3)$, so $\mathrm{dim} (U) = 2$. Furthermore, $\{ x^2, x^3 \}$ are linearly independent in $V$, so take $\{ x^2, x^3 \}$ to be our start up basis.

We must now use the Gram-Scmidt procedure in order to convert this basis to an orthonormal basis. Let $v_1 = x^2$ and $v_2 = x^3$. Let $\{ e_1, e_2 \}$ be our orthonormal basis. Then we have that:

(5)

(6)

Therefore $\{ \sqrt{5}x^2, \sqrt{7} (6x^3 – 5x^2) \}$ is an orthonormal basis of $U$.

We now let $p(x) = P_U(q(x)) = P_U(2 + 3x)$, which is given by the formula:

(7)

\sqrt{5}x^2 +\sqrt{7} (6x^3 – 5x^2) \\ \quad = \sqrt{5}x^2 + \sqrt{7} (6x^3 – 5x^2) = 24x^2 – \frac{203}{10}x^3 \end{align}