In the MAT7381 course (graduate course on regression models), we will talk about optimization, and a classical tool is the so-called conjugate. Given a function \(f:\mathbb{R}^p\to\mathbb{R}\) its conjugate is function \(f^{\star}:\mathbb{R}^p\to\mathbb{R}\) such that \(f^{\star}(\boldsymbol{y})=\max_{\boldsymbol{x}}\lbrace\boldsymbol{x}^\top\boldsymbol{y}-f(\boldsymbol{x})\rbrace\)so, long story short, \(f^{\star}(\boldsymbol{y})\) is the maximum gap between the linear function \(\boldsymbol{x}^\top\boldsymbol{y}\) and \(f(\boldsymbol{x})\).

Just to visualize, consider a simple parabolic function (in dimension 1) \(f(x)=x^2/2\), then \(f^{\star}(\color{blue}{2})\) is the maximum gap between the line \(x\mapsto\color{blue}{2}x\) and function \(f(x)\).

x = seq(-100,100,length=6001) f = function(x) x^2/2 vf = Vectorize(f)(x) fstar = function(y) max(y*x-vf) vfstar = Vectorize(fstar)(x)

We can see it on the figure below.

viz = function(x0=1,YL=NA){ idx=which(abs(x)<=3) par(mfrow=c(1,2)) plot(x[idx],vf[idx],type="l",xlab="",ylab="",col="blue",lwd=2) abline(h=0,col="grey") abline(v=0,col="grey") idx2=which(x0*x>=vf) polygon(c(x[idx2],rev(x[idx2])),c(vf[idx2],rev(x0*x[idx2])),col=rgb(0,1,0,.3),border=NA) abline(a=0,b=x0,col="red") i=which.max(x0*x-vf) segments(x[i],x0*x[i],x[i],f(x[i]),lwd=3,col="red") if(is.na(YL)) YL=range(vfstar[idx]) plot(x[idx],vfstar[idx],type="l",xlab="",ylab="",col="red",lwd=1,ylim=YL) abline(h=0,col="grey") abline(v=0,col="grey") segments(x0,0,x0,fstar(x0),lwd=3,col="red") points(x0,fstar(x0),pch=19,col="red") } viz(1)

or

viz(1.5)

In that case, we can actually compute \(f^{\star}\), since \(f^{\star}(y)=\max_{x}\lbrace xy-f(x)\rbrace=\max_{x}\lbrace xy-x^2/2\rbrace\)The first order condition is here \(x^{\star}=y\) and thus\(f^{\star}(y)=\max_{x}\lbrace xy-x^2/2\rbrace=\lbrace x^{\star}y-(x^{\star})^2/2\rbrace=\lbrace y^2-y^2/2\rbrace=y^2/2\)And actually, that can be related to two results. The first one is to observe that \(f(\boldsymbol{x})=\|\boldsymbol{x}\|_2^2/2\) and in that case \(f^{\star}(\boldsymbol{y})=\|\boldsymbol{y}\|_2^2/2\) from the following general result : if \(f(\boldsymbol{x})=\|\boldsymbol{x}\|_p^p/p\) with \(p>1\), where \(\|\cdot\|_p\) denotes the standard \(\ell_p\) norm, then \(f^{\star}(\boldsymbol{y})=\|\boldsymbol{y}\|_q^q/q\) where\(\frac{1}{p}+\frac{1}{q}=1\)The second one is the conjugate of a quadratic function. More specifically if \(f(\boldsymbol{x})=\boldsymbol{x}^{\top}\boldsymbol{Q}\boldsymbol{x}/2\) for some definite positive matrix \(\boldsymbol{Q}\),  \(f^{\star}(\boldsymbol{y})=\boldsymbol{y}^{\top}\boldsymbol{Q}^{-1}\boldsymbol{y}/2\). In our case, it was a univariate problem with \(\boldsymbol{Q}=1\).

For the conjugate of the \(\ell_p\) norm, we can use the following code to visualize it

p = 3 f = function(x) abs(x)^p/p vf = Vectorize(f)(x) fstar = function(y) max(y*x-vf) vfstar = Vectorize(fstar)(x) viz(1.5)

or

p = 1.1 f = function(x) abs(x)^p/p vf = Vectorize(f)(x) fstar = function(y) max(y*x-vf) vfstar = Vectorize(fstar)(x) viz(1, YL=c(0,10))

Actually, in that case, we almost visualize that if \(f(x)=|x|\) then\(\displaystyle{f^{\star}\left(y\right)={\begin{cases}0,&\left|y\right|\leq 1\\\infty ,&\left|y\right|>1.\end{cases}}}\)

To conclude, another popular case, \(f(x)=\exp(x)\) then\(\){\displaystyle f^{\star}\left(y\right)={\begin{cases}y\log(y)-y,&y>0\\0,&y=0\\\infty ,&y<0.\end{cases}}}[/latex]We can visualize that case below

f = function(x) exp(x) vf = Vectorize(f)(x) fstar = function(y) max(y*x-vf) vfstar = Vectorize(fstar)(x) viz(1,YL=c(-3,3))