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I'm having trouble with the following problem. I am looking at the expression $$ \exp\left(\frac{1}{\lambda}\int_0^t \left(c-\tfrac{1}{2}g(s)\right)^2 \mathrm{d} s\right)\quad (*), $$ where $t>0$ is fixed and $g$ is an integrable bounded function and $(c,\lambda)\in\mathbb{R}\times (0,\infty)$. Furthermore we let $$ l=\inf_{u\in\mathbb{R}} \tfrac{1}{2}g(u),\quad r=\sup_{u\in\mathbb{R}}\tfrac{1}{2}g(u) - l. $$ Now I want to find an upper bound on the expression in $(*)$ using $l$ and $r$. This is what I have come up with: $c-r-l\leq c-\tfrac{1}{2}g(s)\leq c-l$ and hence $$ \int_0^t\left(c-\tfrac{1}{2}g(s)\right)^2 \mathrm{d} s\leq t \max ((c-l)^2,(c-r-l)^2). $$ Supposing this bound is correct, I also want to minimize the upper bound in terms of $(c,\lambda)\in\mathbb{R}\times (0,\infty)$, i.e. minimize $$ \exp\left(\frac{t}{\lambda}\max ((c-l)^2,(c-r-l)^2) \right),\quad (c,\lambda)\in\mathbb{R}\times (0,\infty). $$ What bothers me, is the $\max$ in the exponent of the last expression which makes me unable to minimize the function. Is there a better bound than the one I came up with? Thanks in advance.

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$$ \min\left\{\max\left\{(c-\ell)^2,(c-r-\ell)^2\right\};c\in\mathbb R\right\}=\tfrac14r^2 $$ $$ \inf\left\{\frac{t}\lambda\max\left\{(c-\ell)^2,(c-r-\ell)^2\right\};c\in\mathbb R,\lambda\gt0\right\}=0\qquad (t\gt0) $$ The minimum in the first formula is attained when the two arguments in the maximum coincide, that is, at $c=\ell+\frac12r$. The infimum in the second formula is attained when $\lambda\to+\infty$.

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  • $\begingroup$ Thanks for your fast reply! Of course I forgot to mention that I am also interested in for what values of $c$ and $\lambda$ the minimum is attained. Is that something you could elaborate on? $\endgroup$ – Stefan Hansen May 28 '12 at 9:28
  • $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did May 28 '12 at 9:37
  • $\begingroup$ Again, thanks alot. $\endgroup$ – Stefan Hansen May 28 '12 at 9:39

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