let let $x(t)$ and $y(t)$ be two state variables belonging to $\mathbb{R}^{n\times 1}$. Assume that $\Vert y(t)\Vert <y_{max}$ for any $t$ with $y_{max} >0$. Then, how can I obtain the following inequality

$$\int_{t_1}^{t_1+T_0}{\Vert y^T(\tau)(x(t_1)-x(\tau))\Vert d\tau} \le y_{max}T_0 \sup_{\tau\in[t_1,t_1+T_0]}{\Vert x(t_1)-x(\tau)\Vert}$$ where $\Vert\cdot \Vert$ is the euclidean norm.

I guess that using triangular inequality one has that $\Vert y^T(\tau)(x(t_1)-x(\tau))\Vert \le \Vert y^T(\tau)\Vert \Vert(x(t_1)-x(\tau))\Vert$ yielding

$$\int_{t_1}^{t_1+T_0}{\Vert y^T(\tau)(x(t_1)-x(\tau))\Vert d\tau} \le y_{max}\int_{t_1}^{t_1+T_0}{\Vert(x(t_1)-x(\tau))\Vert d\tau}$$

but then my problem is in finding the upper-bound of the integral in terms of the supremum of $\Vert(x(t_1)-x(\tau))\Vert$ on the integration interval $[t_1,t_1+T_0]$. Can some one give me a hand to understand this. Thanks


Suppose we are looking at the value of the integrandum. This value is nonnegative since it is given by the euclidean distance, and by defenition of the supremum it is valid that $$\forall \tau \in [t_1, t_1+T_0] : \Vert x(t_1) - x(\tau) \Vert \leq \sup_{\tau\in[t_1,t_1+T_0]}{\Vert x(t_1)-x(\tau)\Vert} = \sup_{\mu\in[t_1,t_1+T_0]}{\Vert x(t_1)-x(\mu)\Vert}$$ (this last renaming step is made to avoid confusion) thus, we can use this in the integral you found:

$$y_{max}\int_{t_1}^{t_1+T_0}{\Vert(x(t_1)-x(\tau))\Vert d\tau} \leq y_{max}\int_{t_1}^{t_1+T_0}{\sup_{\mu\in[t_1,t_1+T_0]}{\Vert x(t_1)-x(\mu)\Vert}d\tau} $$

the last step is to recognise that this supremum is independant of the value of $d\tau$, so we can just put it outside. if we then calculate $$\int_{t_1}^{t_1+T_0}{d\tau} = T_0 $$ we see the the answer as needed arise (rename again):

$$\int_{t_1}^{t_1+T_0}{\Vert y^T(\tau)(x(t_1)-x(\tau))\Vert d\tau} \le y_{max}T_0 \sup_{\tau\in[t_1,t_1+T_0]}{\Vert x(t_1)-x(\tau)\Vert}$$

  • $\begingroup$ Alex, many thanks! $\endgroup$ – Dan Apr 19 '17 at 18:22

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