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I am having trouble with this proof. I am given the following 2 definitions:

1) A function $f$ is coercive if $\lim_{||x|| \rightarrow \infty} f(x) = \infty$

2) A $C^2$ function $f$ is strongly convex if there exists a constant $c_0 > 0$ such that: $(x - y)^T (\nabla f(x) - \nabla f(y)) \geq c_0 ||x - y||^2 \hspace{5mm}$ $\forall x,y \in \mathbb{R}^n$

The question is to show that if $f$ is strongly convex then it is coercive.

I am only allowed to use basic theorems such as Taylor expansion, triangle inequality etc. However, I can use the fact that a function is coercive $\iff$ all its level sets are compact.

My instinct is that the answer can be obtained by a doing a Taylor expansion and manipulating the result, but I've been stuck for days using this approach. Any help would be greatly appreciated.

r.emp
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3 Answers3

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Lemma 1: Let $f:\mathbb R^n\to \mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $\forall x,y\in \mathbb R^n, (\nabla f(y)-\nabla f(x))^T(y-x)\geq 0$.

Lemma 2: Let $f:\mathbb R^n\to \mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $\forall x,y\in \mathbb R^n, f(y)\geq f(x)+\nabla f(x)^T(y-x)$

Let us prove first that your definition of "strongly convex" implies convexity of $\displaystyle g:x\mapsto f(x)-\frac{c_0}2\|x\|^2$

Indeed, $\begin{aligned}[t](\nabla g(y)-\nabla g(x))^T(y-x)&=(\nabla f(y)-c_0y - (\nabla f(x)-c_0x))^T(y-x)\\ &=(\nabla f(y)-\nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \\ &= (\nabla f(y)-\nabla f(x))^T(y-x)-c_0\|y-x\|^2 \\ &\geq 0 \quad \text{ with your definition} \end{aligned}$

By Lemma 1, $g$ is convex.

By Lemma 2, $$\begin{aligned}[t] g(y)&\geq g(x)+\nabla g(x)^T(y-x) \\ f(y)-\frac{c_0}2\|y\|^2 &\geq f(x)-\frac{c_0}2\|x\|^2 + (\nabla f(x)-c_0x)^T(y-x) \end{aligned}$$ Let $\alpha =\nabla f(x)-c_0x$. Then $$f(y)\geq \frac{c_0}2\|y\|^2+ \alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.

By Cauchy-Schwarz, $\alpha^Ty \geq -\|\alpha\|\|y\|$, hence $$f(y)\geq \frac{c_0}2\|y\|^2 -\|\alpha\|\|y\|+K_x$$

The right-hand side goes to $\infty$ as $\|y\|\to \infty$ and we're done.

Gabriel Romon
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The proof can be found, for example, in the notes of John Gubner (see Corollary 3.2).

You can also take a look at the proof of Corollary 11.17 in Convex Analysis and Monotone Operator Theory in Hilbert Spaces, by Bauschke and Combettes.

Plop
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maoqunqun
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Another route would be by using the alternative characterizations of strong convexity.

Def.: A function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is strongly convex, if $\exists \mu > 0$ so that $\forall x, y \in \mathbb{R^n}$ and $\lambda \in [\ 0, 1 ]\ $, $f(\lambda x + (1-\lambda) y) + \mu \lambda (1-\lambda) ||x -y||^2 \leq \lambda f(x) + (1-\lambda f(y).$ Further:

Lemma: A continuously differentiable $f$ is strongly convex iff $ \exists \mu >0$, so that $\forall x, y \in \mathbb{R}^n, \lambda \in [\ 0, 1 ]\ $, $ \nabla f(x)^T(y-x) + \mu ||x-y||^2 \leq f(y) - f(x).$

With the second characterisation, which can be obtained for a strongly convex function in a very similar way as the characterisation for convex $f$ from Lemma 2 in the first answer, we can set $x=0$ and obtain

$f(y) \geq \nabla f(0)^Ty + \mu ||y||^2 + f(0) \geq -||\nabla f(0)||.||y|| + \mu ||y||^2 + f(0), \forall y \in \mathbb{R}^n$ with Cauchy- Schwarz and clearly $f(y) \rightarrow \infty$, if $||y|| \rightarrow \infty.$

Len
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