I was studying inner products from the book of Linear Algebra by Friedberg when I came across this example
The verification of the first three conditions are clear to me. But I couldn't get the last one i.e. $\langle f,f\rangle > 0$ and why $f^2$ is bounded in some subinterval of $[0,1]$? And one more thing, why does this result in the truth of the last condition, i.e. why is the integral value greater than $0$? Any explanation will be highly beneficial.
It is clear that $f^2$ is nonnegative.
Since $f\ne0$, for some $x_0 \in [0,1]$, $f(x_0)\ne0$.
By continuity, $f(x_0) \ne0$ in a (small) interval around $x$.
(The above sentence is the meaning of "bounded away from zero".)
Integrating $f^2$ in that interval gives a (strictly) positive value.
Explicitly, there exists some $\delta > 0$ such that for any $x \in (x_0-\delta, x_0+\delta)$ (if $x_0=0$ or $1$ this is one-sided)
$$|f(x)|>\left|\frac{f(x_0)}2\right|$$
This leads to:
$$f^2(x) > \frac {f^2(x_0)}4 > 0$$
and therefore:
$$\int_0^1 f^2(x)dx\ge\int_{x_0-\delta}^{x_0+\delta}f^2(x)dx>\int_{x_0-\delta}^{x_0+\delta} 0 dx=0$$