Let $f:\mathbb{R}^n\to\mathbb{R}$ .Prove $f$ is convex iff for any $x\in\mathbb{R}^n$ and $d\neq 0$ $g_{x,d}(t)=f(x+td)$ is convex(where g is one-dimensional).
The way that we assume that $f$ is convex and proving $g$ is I have managed to do but the other direction is tried some stuff but didn't really get any close.
Any hint please?
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Arctic Char
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convxy
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1What is your definition of convex? – angryavian Nov 10 '20 at 16:17
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$f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$ and $\lambda\in[0,1]$ – convxy Nov 10 '20 at 16:19
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1Choose $d=y-x$ and apply convexity of $g_{x,d}$ on the interval $[0,1]$. – angryavian Nov 10 '20 at 16:20
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Ok, I'll try but what is the intuition behind it – convxy Nov 10 '20 at 16:21
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@ronkurman the intuition is that inside $f$, we end up having something resembling a convex combination: $x+t(y-x)=ty+(1-t)x$ (particularly useful when $t\in[0,1]$) – Zim Nov 10 '20 at 18:49