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I am having some problem determining all the cases for finding the conjugate of a function.

Find the conjugate of:

(i) $f(x) = e^x$ , $x ∈ \mathbb {R}$

For this, I determined that $f^* = x^*ln(x^*) - x^*$ and I thought this was the answer but I had to consider other cases which I am not sure how to determine.

(ii) $f(x) = \rho ||x||_1$

For this, I have to consider the cases when $||z||_\infty \le \rho$ and $||z||_\infty \gt \rho$ but I don't know what is the process to think of these cases. Any help is appreciated

eun ji
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1 Answers1

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This really is a case of grinding through.

(i) $f^*(\alpha) = \sup_x \alpha x - e^x$. Note that for $\alpha <0$, if $x<0$ we have $\alpha x -e^x > \alpha x -1$ and so $f^*(\alpha) = \infty$. If $\alpha = 0$, then choose large negative $x$ we see that $f^*(0) = 0$. Finally, as you have done in the question, for $\alpha>0$ we have $f^*(\alpha) = \alpha( \log \alpha -1)$.

Hence $f^*(\alpha) = \begin{cases}\infty, & \alpha < 0 \\0, & \alpha =0 \\ \alpha( \log \alpha -1), & \alpha >0 \end {cases}$.

(ii) $f^*(\alpha) = \sup_x \sum_k \alpha_k x_k - \rho |x_k|$.

It is straightforward to see that if $\| \alpha\|_\infty > \rho$ then $f^*(\alpha) =\infty$.

If $\| \alpha\|_\infty \le \rho$, then $f^*(\alpha) \le \sup_x \sum_k (|\alpha_k| - \rho )|x_k| \le 0$, so $f^*(\alpha) = 0$.

Hence $f^*(\alpha) = \begin{cases} 0,& \|\alpha\|_\infty \le \rho \\ \infty, & \|\alpha\|_\infty > \rho\end {cases}$.

copper.hat
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  • may I know when we do not need to split the conjugate to cases? @copper.hat – eun ji Mar 13 '20 at 06:24
  • Hmm, there is no real recipe for when you do & don't. Look for directions in which the function is linear maybe? – copper.hat Mar 13 '20 at 06:32
  • For example, the conjugate of a quadratic function is just one case. So there's normally more than one case when the function is linear? @copper.hat – eun ji Mar 13 '20 at 06:57
  • There is no recipe really, but you will see some patterns, much like one learns patterns with integration. For (i) above. for example, I do a rough plot of $-e^x$ and then draw various lines at different angles to see if a shifted version of the line can be an upper bound for $-e^x$. For example, if the slope is negative, then it is clear that the $\sup$ must be $+\infty$. I had to play with the formula to get (ii). – copper.hat Mar 13 '20 at 16:33