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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.

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

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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*}

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