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Let $f: \mathbb{R^3} \rightarrow \mathbb{R}, f(x) = x\cdot(x-a)$ where $\cdot$ is the dot product. I want to show that $f$ is convex directly from the definition of convex functions, that is $\forall x, y \in \mathbb{R^3}, \theta \in [0, 1], f(\theta x+(1 - \theta)y) \leq \theta f(x) + (1 - \theta) f(y).$ After I plug in the function in to the definition of convexity and simplifying, I get $(\theta x + (1 - \theta) y) \cdot (\theta x + (1 - \theta) y) \leq \theta x\cdot x + (1 - \theta)y \cdot y$, and I can't get any further. How can we conclude this expression is true? Or I did something wrong during the simplification?

Charlie
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    Hint: First show that $x.x=|x|^{2}$ is convex. – Kavi Rama Murthy Oct 07 '21 at 23:26
  • @KaviRamaMurthy How does this help to show $x \cdot (x - a)$ is convex? – Charlie Oct 07 '21 at 23:28
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    Hint 2: $x\cdot(x-a) = | x |^2 - a.x $ :) – Red shoes Oct 07 '21 at 23:30
  • $x.(x-a)=x.x-x.a$ – Kavi Rama Murthy Oct 07 '21 at 23:30
  • @KaviRamaMurthy Keep in mind that I am asking this question out of ignorance. I was under the impression that if $f(x)$ is any function where $f''(x)$ is defined and always positive on an interval $(a,b)$, then $f(x)$ is convex (i.e. curving upwards) on that interval. Is this correct? If so, then isn't the resolution to the problem simply that if $f(x) = x(x-a)$ then $f''(x) = 2$? – user2661923 Oct 08 '21 at 02:26
  • @KaviRamaMurthy It just occurred to me that I am confusing $x \times (x-a)$ with $x \cdot (x-a)$. If this confusion is the only stumbling block, re the previous comment, can the analysis in the previous comment be somehow used to apply to $f(x) = x \cdot (x-a)$? – user2661923 Oct 08 '21 at 02:31

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You almost got there. $(\theta x+(1-\theta )y)\cdot (\theta x+(1-\theta )y)=\theta^2||x||^2+2\theta(1-\theta)||x||\cdot ||y||+(1-\theta)^2||y||^2 $. Now since $2||x||\cdot ||y||\le ||x||^2+||y||^2$, your result is right there.

toronto hrb
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