There is a proof in my textbook where I am a little bit unsure about a small detail. It would be great if someone could clarify it for me.
We are supposed to prove positivity of the $L^2$ inner product on the space of continuous functions on $[a,b]$. The book does this as follows:
We wish to show that the $L^2$ inner product on the interval $[a,b]$ satisfies the positivity axiom; this means we must show that if $0 = \langle f,f \rangle = \int_{a}^{b} |f(t)|^2 dt$, then $f(t) = 0$ is zero for all $a < t < b$. Suppose by contradiction that there is a $t_0$ with $f(t_0) \neq 0$. Let $\displaystyle\epsilon = \frac{|f(t_0)|}{2} > 0$ in the definition of continuity; which states that there is a $\delta > 0$ such that it $|t - t_0| < \delta$, then $\displaystyle|f(t) - f(t_0)| < \epsilon = \frac{|f(t_0)|}{2} > 0$. This implies $\displaystyle|f(t)| > \frac{|f(t_0)|}{2}$ for $|t - t_0| < \delta$. Assuming $\delta$ is chosen small enough so that the interval $[t_0 - \delta, t_0 + \delta]$ is contained in $[a,b]$, then
$\int_{a}^{b}|f(t)|^2 dt \geq \int_{t_0 - \delta}^{t_0 + \delta} |f(t)|^2 dt \geq \frac{|f(t_0)|^2}{4} [2 \delta] > 0$
This shows that if $f(t_0) \neq 0$, then $\langle f,f \rangle > 0$. Therefore, we conclude that if $\langle f,f \rangle = 0$, then $f(t_0) = 0$ for all $a < t_0 < b$.
OK, so the only thing I don't fully understand here is that if we have shown that $|f(t)|$ is greater than $\displaystyle\frac{|f(t_0)|}{2}$ for $|t - t_0| < \delta$, then why, when we look at the integrals later do we write: $\displaystyle\int_{t_0 - \delta}^{t_0 + \delta} |f(t)|^2 dt$ is greater or equal to $\displaystyle\frac{|f(t_0)|^2}{4} [2 \delta]$. If someone could explain to me why we use "greater or equal to" in this last step, then I would be extremely grateful!