Convex functions

Show that the following functions from \(\mathbb{R}^{n}\) to \((-∞,∞]\) are convex :

\(\hspace{1cm}(a)\hspace{0,5cm} f_{1}\left( x\right) =\ln \left( e^{x_{1}}+\ldots+e^{x_{n}}\right)\)

\(\hspace{1cm}(b)\hspace{0,5cm} f_{2}\left( x\right) =\left\Vert x\right\Vert^{p},\) with \(p\geq 1\)

\(\hspace{1cm}(c)\hspace{0,5cm}f_{3}\left( x\right) =\frac{1}{f\left( x\right) },\) where \(f\) is concave and \(0<f(x)<\infty\) for all \(x\)

\((a)\) We show that the Hessian of \(f_{1}\) is positive semidefinite at all \(x∈\mathbb{R}^{n}\). Let \(β(x)=e^{x₁}+...+e^{x_{n}}\). Then a straightforward calculation yields

\(\hspace{1,5cm}z^{T }\nabla ^{2}f_{2}\left( x\right) z=\frac{1}{\beta \left( x\right)^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}e^{\left( x_{i}+x_{j}\right) }\left(z_{i}-z_{j}\right) ^{2}\geq 0,\forall z\in \mathbb{R}^{n},\)

so \(f_{1}\) is convex.

\((b)\) The function\( f_{2}(x)=‖x‖^{p}\) can be viewed as a composition \( g(f(x))\) of the scalar function \(g(t)=t^{p}\) with \(p≥1\) and the function \(f(x)=‖x‖\). In this case, \(g\) is convex and monotonically increasing over the nonnegative axis, the set of values that f can take, while f is convex over \(\mathbb{R}^{n}\) (since any vector norm is convex ). Using the properties of the power function, we deduce that \(f_2\) is convex.

\((c)\) The function \(f_{3}(x)=\frac{1}{f(x)}\) can be viewed as a composition \(g(h(x))\) of the function \(g(t)=-\frac{1}{t}\) for \( t<0\) and the function \( h(x)=-f(x)\) for \(x∈\mathbb{R}^{n}\). In this case \(g\) is convex and monotically increasing in the set \({t<0}\), while h is convex over \(\mathbb{R}^{n}\). Using the properties of the inverse function, it follows that the function \(f_{3}=\frac{1}{f(x)}\) is convex over \(\mathbb{R}^{n}\).