Question

In C [a,b], define the product

$<f,g>\mathbf{=}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}\left(t\right)\mathbf{g}\left(t\right)\mathbf{dt}$

Show that this product satisfies the property$<f,f>\mathbf{>}\mathbf{0}$ for all nonzero f.

Short answer

It shows that .

Step-by-step solution

Consider the value of aσ>0 .

Since F is not identically zero, so let $f\left(x_0\right)\ne 0$. As f is continuous, so for $\in =\frac{\left|f\left(x_0\right)\right|}{2}$we will able to find such that

$x\in \left(x_0-{\delta }_1{x}_0+\delta \right)\cap \left[a,b\right]\Rightarrow \left|f\left(x\right)-f\left(x_0\right)\right|<\frac{\left|f\left(x_0\right)\right|}{2}$

Thus,

$$\begin{gathered} x\in \left(x_0-{\delta }_1{x}_0+\delta \right)\cap \left[a,b\right]\Rightarrow \left|f\left(x\right)-f\left(x_0\right)\right|<\frac{\left|f\left(x_0\right)\right|}{2} \\ \Rightarrow \left|\left|f\left(x\right)-f\left(x_0\right)\right|\right|\frac{\left|f\left(x_0\right)\right|}{2}\left|\left|a\right|-\left|b\right|\right|\le \left|a-b\right| \\ \Rightarrow \frac{\left|f\left(x_0\right)\right|}{2}<\left|f\left(x\right)\right|<\frac{3\left|f\left(x_0\right)\right|}{2} \end{gathered}$$

Thus,

$\left|f\left(x\right)\right|<\frac{\left|f\left(x_0\right)\right|}{2}$

Now, observe that,

$$\begin{gathered} >\underset{\left(x_0-\delta ,x_0+\delta \right)\cap \left[a,b\right]}{\int }{\left(\frac{\left|f\left(x_0\right)\right|}{2}\right)}^{2}dt \\ ={\left(\frac{\left|f\left(x_0\right)\right|}{2}\right)}^2\times 2\delta \\ =\delta \frac{{\left|f\left(x_0\right)\right|}^2}{2} \\ >0 \end{gathered}$$

Hence, .