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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!

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@Matt He's only dealing with continuous functions. –  Thomas Andrews Sep 7 '12 at 13:16
Greater than implies greater than or equal to. Doesn't hurt to say $a>b$ implies $a\geq b$. –  Thomas Andrews Sep 7 '12 at 13:17
Since he's talking about continuous functions, he'll find $f=0$ on all of $[a,b]$. –  Hagen von Eitzen Sep 7 '12 at 13:18
@ThomasAndrews: Thanks a lot. Yeah, I just thought it was a bit odd that they changed notation like that. –  Kristian Sep 7 '12 at 13:19
But presumably they haven't gotten there - this theorem essentialy is proving that statement from $\epsilon-\delta$ point of view. @Kristian –  Thomas Andrews Sep 7 '12 at 13:19

1 Answer 1

up vote 1 down vote accepted

Well you chose your $\delta$ so that $f|_{(t_0-\delta,t_0+\delta)} \gt f(t_0)/2 > 0$, but you're ending up integrating over the closed interval not the open one. At the end points of the interval we can't get a strict inequality only a weak one namely $f|_{[t_0-\delta, t_0+\delta]} \geq f(t_0)/2$ so you're just giving a very weak approximation to the integral from below: namely you're squaring both sides and integrating. In fact it will have to be a strict inequality(since $f > c$ at some point in this interval, apply this theorem to again to $f-c$).

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Thanks a lot for clarifying this! Really appreciate it. –  Kristian Sep 7 '12 at 13:51

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