# Convex function properties

If $$f(\textbf{x}): R^n \rightarrow R$$ is a convex function on $$S \subseteq R^n$$, how can we show that $$f(t) = f(\textbf{x} + t\Delta \textbf{x})$$ is a convex function on $$\{t \in R : t>0\}$$? We assume that $$\textbf{x} + t\Delta \textbf{x} \in S$$. I tried to use the definition of a convex function but could not get anywhere.

• What happened when you tried to use the definition of a convex function? – littleO May 16 at 0:56
• In general, if $f$ is convex and $g$ is affine it is almost immediate that $f \circ g$ is convex. – copper.hat May 16 at 3:35

Consider the injective affine map $$T:\Bbb R^2\to \Bbb R^{n+1}\\ T(t,y)=(x+t\Delta x,y)$$
Then, $$T(\operatorname{epi}(f(x+\bullet\cdot \Delta x))=\operatorname{im}T\cap \operatorname{epi}f$$.
Call $$G:\operatorname{im}T\to \Bbb R^2$$ the affine map such that $$G\circ T=\operatorname{id}$$. Then, $$\operatorname{epi}(f(x+\bullet\cdot \Delta x))=G[\operatorname{im}T\cap\operatorname{epi}f]$$ and image by an affine function of a convex set is convex.
A fortiori, $$\operatorname{epi}(f(x+\bullet\cdot \Delta x))\cap ((0,\infty)\times\Bbb R)$$ will be convex.
Take $$t_{1},t_{2} > 0$$. We will show that $$\forall h \in \mathbb{R}, 0 < h < 1$$, we have $$f(ht_{1} + (1-h)t_{2}) \leq hf(t_{1}) + (1-h)f(t_{2})$$.
Observe that : \begin{align*} f(ht_{1} + (1-h)t_{2}) &=f(x + (ht_{1} + (1-h)t_{2})\Delta x)\\ &= f(h(x + t_{1}\Delta x) + (1-h)(x + t_{2}\Delta x)) \\ &\leq hf(t_{1}) + (1-h)f(t_{2}) \text{(by convexity of }f(x)\text{)} \end{align*}