Regarding Schwartz inequality and integration bounds

[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:
$\left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq s\int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr,$
instead of:
$\left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq \int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr.$

 

[S.G. Janssens]

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[p4wp4w]

You may want to type this in $\LaTeX$ to make it readable. Here is a guide: https://www.physicsforums.com/help/latexhelp/

done!

 

[Samy_A]

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:
$\left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq s\int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr,$
instead of:
$\left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq \int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr.$

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

 

[p4wp4w]

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

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.