Let function $f : \mathbb{R}^n \to \mathbb{R}$ with $\|\cdot\|$ is Euclidean norm then we have the following definitions:
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A function $f$ is coercive if $\lim\limits_{\|x\|\rightarrow \infty}f(x) = \infty$.
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A function $f$ is convex if for all $x,y\in \mathbb{R}^n$ and $\lambda \in [0;1]$
$$f((1-\lambda)x + \lambda y)\leq (1-\lambda)f(x) + \lambda f(y)$$ -
A function $f$ is strongly convex if there is exists a constant $\kappa >0$ such that for all $x,y\in\mathbb{R}^n$ and $\lambda \in [0;1]$
$$f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y)-\dfrac{\kappa}{2} \|x-y\|^2$$
Problem: Prove that if $f$ is strongly convex then $f$ is coercive.
Note: There is another post about this kind of problem but the definition of strongly convex function on that post used a $C^2$ function $f$. I think that the following lemmas maybe useful for completing the proof.
Lemma 1: If $f$ is strongly convex function then $f$ is convex.
Lemma 2: $f$ is strongly convex function with some $\kappa>0$ iff $f – \dfrac{\kappa}{2}\|\cdot\|$ is convex function.
Best Answer
There might be a typo in the question, the condition for strong convexity should be $$f((1-\lambda)x+\lambda y)\leq(1-\lambda)f(x)+\lambda f(y)-\frac\kappa2{\color{red}{\lambda(1-\lambda)}}\|x-y\|^2.\tag{SC}$$ If there is no $\lambda(1-\lambda)$, letting $y=0,\lambda=0$ will lead to $\|x\|\leq0$ for any $x\neq0$, which is false.
Assume $f(0)=0$, let $y=0$, $\lambda=1/2$, by (SC) we have $$f(x)\geq\frac\kappa4\|x\|^2+2f(x/2).$$ Since $f$ is convex, we have $f(cx)\leq cf(x)$ for $c\in[0,1]$. Suppose $\|x\|>2$ and choose $c=2/\|x\|$, it follows $$2f(x/2)\geq\|x\|f(x/\|x\|).$$ Let $u_x=x/\|x\|$ be the unit vector in the direction of $x$, then $$f(x)\geq\frac\kappa4\|x\|^2+f(u_x)\|x\|.$$ As $\|x\|\to\infty$ along the same direction, $u_x$ stays the same and therefore $f(x)\to\infty$ due to the quadratic term. This shows $f$ blows up along all directions, which is coercivity. For $f$ with $f(0)\neq0$, simply use the above arguments for $f-f(0)$.