Question

In the space of the polynomials of degree, we define the inner product

$<f,g>\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(f\left(0\right)+g\left(0\right)+f\left(1\right)+g\left(1\right)\right)$

Find an orthonormal basis for this inner product space.

Short answer

The orthonormal basis of$P_1$ is$\left\{1,2t-1\right\}$ .

Step-by-step solution

Determine the Gram-Schmidt process.

Consider a basis of a subspace Vof ${\mathit{R}}^{\mathbf{n}}$ for j = 2,...,m we resolve the vector ${\overrightarrow{\mathbf{v}}}_{\mathbf{j}}$into its components parallel and perpendicular to the span of the preceding vectors localid="1660108333006" ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}}{\overrightarrow{\mathbf{v}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$,

Then${\overrightarrow{\mathit{u}}}_{\mathit{1}}\mathit{=}\frac{\mathit{1}}{||{\overrightarrow{v}}_1||}{\overrightarrow{\mathit{v}}}_{\mathit{1}}\mathit{,}{\overrightarrow{\mathit{u}}}_{\mathit{2}}\mathit{=}\frac{\mathit{1}}{||{{\overrightarrow{v}}_2}^{\perp }||}{{\overrightarrow{\mathit{v}}}_{\mathit{2}}}^{\mathit{\perp }}\mathit{,}\mathit{.}\mathit{.}\mathit{.}\mathit{,}{\overrightarrow{\mathit{u}}}_{\mathit{j}}\mathit{=}\frac{\mathit{1}}{||{{\overrightarrow{v}}_j}^{\perp }||}{{\overrightarrow{\mathit{v}}}_{\mathit{j}}}^{\mathit{\perp }}\mathit{,}\mathit{.}\mathit{.}\mathit{.}\mathit{,}{\overrightarrow{\mathit{u}}}_{\mathit{m}}\mathit{=}\frac{\mathit{1}}{||{{\overrightarrow{v}}_m}^{\perp }||}{\overrightarrow{\mathit{v}}}_{\mathit{m}}$

It will use the Gram-Schmidt process to get an orthonormal basis.

Let$f_1\left(t\right)=1,f_2\left(t\right)=t$for find the first element of orthonormal basis is,

${||f||}^2=\frac{1}{2}\left(f\left(0\right)+f\left(0\right)+f\left(1\right)+f\left(1\right)\right)=1$

Therefore take

Now, observe the following

Further solve the above expression,

$$\begin{gathered} g_2\left(t\right)=\frac{t-1/2}{1/2} \\ =\left(2t-1\right) \end{gathered}$$

Hence, an orthonormal basis of$P_1$ is$\left\{1,2t-1\right\}$