Let $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=e^x$. Determine $f^*(y)$.
I try to use some inequalities to get supremum but it is impossible. Seemingly, I must consider some cases of $x$.
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1What exactly do you mean by $f^*$? – Abel May 22 '13 at 13:09
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I assume he means $f^*(y)\triangleq \sup_x\langle y,x\rangle - f(x)$, given the title. – Michael Grant May 22 '13 at 13:14
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@Abel: It is like what Michael says. $f^*(y)=\sup_{x\in\mathbb{R}}\left{xy-f(x)\right}$ – anhkhoavo1210 May 22 '13 at 13:16
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$$f^*(y)=\textstyle\sup_x \langle y, x \rangle - e^x.$$ Since the supremum involves a smooth function we can examine the derivative: $$\frac{d}{dx}(yx-e^x) = y-e^x = 0\quad\Longrightarrow\quad x=\log y~~\text{if}~~y>0.$$ So if $y>0$, at least, $$f^*(y) = y\log y - y.$$ If $y=0$, then the supremum is $0$; if $y<0$, then the supremum is $+\infty$. (In both cases, let $x\rightarrow-\infty$.) So the final answer is $$f^*(y)=\begin{cases} y\log y - y & y > 0 \\ 0 & y = 0 \\ +\infty & y < 0 \end{cases}$$ If you don't like extended-valued functions just drop the $y<0$ case and state that $$\mathop{\textrm{dom}}(f^*) = \mathbb{R}_+.$$
Michael Grant
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why using derivative for y>0, I can't understand. Besides, what will happen if $x\to+\infty$ – hung tran May 22 '13 at 15:39
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$x$ is the independent variable inside the supremum. The proper value of $x$ is the one that $yx-e^x$. There's no way that it can approach infinity, because $x\rightarrow+\infty$ means $yx-e^x\rightarrow-\infty$. – Michael Grant May 22 '13 at 15:44
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Hung Tran asks: Could you tell me more detail about the case $y>0$, why computing $f^*$ by finding extreme point of $x$? – anhkhoavo1210 May 22 '13 at 16:11
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Because it is necessary to compute $\sup_x yx-e^x$. How else would you propose to solve it? – Michael Grant May 22 '13 at 16:52