# Regarding Schwartz inequality and integration bounds

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1. Feb 2, 2016

### p4wp4w

Based on Schwartz inequality, I am trying to figure out why there
should/can be the "s" variable which is the lower bound of the
integration in the RHS of the following inequality:
$\left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq s\int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr,$
$\left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq \int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr.$

Last edited: Feb 2, 2016
2. Feb 2, 2016

3. Feb 2, 2016

4. Feb 2, 2016

### Samy_A

Apply the Cauchy-Schwarz inequality to the functions$f,\ g$ defined by $f(r)=A(t+r)Z(t+r)$ and $g(r)=1$.

5. Feb 2, 2016

### p4wp4w

Sure. My confusion was because of the example in my mind that s can be a very small value which then the upper bound will be also very small and as a result, physically conservatism but again mathematically correct. Thank you very much.