A quick question from a multiple choice test I am preparing for:
Is the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ with $f(x_1,x_2) = 4x_1^2+x_2^2$ uniformly convex?
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3nondatur
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Sorry I made a mistake I am working on an edit. – 3nondatur Jun 26 '19 at 17:35
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I finished my edit, it should be alright now. – 3nondatur Jun 26 '19 at 17:41
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1I get hat the left hand side is $3(x_1-y_1)^2 + |x-y|^2$ which is the same as you just with $+x_2^2$ and $-2x_2y_2$. Check your signs. – Winther Jun 26 '19 at 17:44
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You are right I corrected the signs. – 3nondatur Jun 26 '19 at 17:52
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1The conclusion is still wrong. Is the left hand side negative for this choice of $x$ and $y$? – Winther Jun 26 '19 at 21:27
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Ahhhh, no it is not... – 3nondatur Jun 26 '19 at 21:46
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Can I find such a $\mu$ at all? – 3nondatur Jun 26 '19 at 22:01
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AHHH, we can use a different criterion that makes it much much easier. I edited my post. – 3nondatur Jun 26 '19 at 23:09
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1You already had the answer if you just factored the left hand side of what you got. A hint has already been given above. – Winther Jun 26 '19 at 23:24
1 Answers
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The key is to remember the fact that
$f$ is uniformly convex (on $X$) iff there is a $\mu > 0$ such that $$d^T \nabla^2(y)d \ge \mu \|d\|^2$$ for all $x \in X$ and all $d \in \mathbb{R}^n$.
This is easy to check.
3nondatur
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