For a space with scalar product (dot product) $C[0,1]$, we have two functions$f(t):t\to t^{2}$ and $g(t): t\to t^3$. Here we have $<f,g> = 1/6$, $||f^2||= 1/5$, and $||g^2||=1/7$.
My question is: How does (in this case my textbook) get these values? Am I correct in thinking that you have to use this formula: $<f,g>:\equiv \displaystyle\int_a^b \! f(t)g(t) \, \mathrm{d}t$ ?
I tried it with this formula, however I did not succeed... Anyone got any ideas?
On $\mathcal{C}\big([0,1],\mathbb{R}\big)$, the inner product is :
$$ \left\langle \cdot,\cdot \right\rangle \, : \, (f,g) \, \longmapsto \; \int_{0}^{1} f(t)g(t) \, dt $$
Now, let $f_{1} \, : \, t \in [0,1] \, \longmapsto \, t^{2}$ and $f_{2} \, : \, t \in [0,1] \, \longmapsto \, t^3$. Then :
$$ \begin{align*} \left\langle f_1,f_2 \right\rangle &= {} \int_{0}^{1} t^{2}t^{3} \, dt \\[2mm] &= \int_{0}^{1} t^5 \, dt \\[2mm] &= \Big[ \frac{1}{6}t^{6} \Big]_{t=0}^{t=1} \\[2mm] &= \frac{1}{6}. \end{align*} $$
Then, $\Vert f_{1} \Vert^2 = \left\langle f_1,f_1 \right\rangle$ and $\Vert f_2 \Vert^2 = \left\langle f_2,f_2 \right\rangle$.