Any example of strongly convex functions whose gradients are Lipschitz continuous in $\mathbb{R}^N$ Let $f:\mathbb{R}^N\to\mathbb{R}$ be strongly convex and its gradient is Lipschitz continuous, i.e., for some $l>0$ and $L>0$ we have $$f(y)\geq f(x)+\nabla f(x)^T(y-x)+\frac{l}{2}||y-x||^2,$$ and $$||\nabla f(y)-\nabla f(x)||\leq L||y-x||$$ for any $x,y\in\mathbb{R}^N$.
For such a function, the only example I can come up with  is the quadratic function $$f(x)=x^T A x +B x + C$$ where $A$ is positive definite. I wonder if there is any other example? Thanks.
Note: the strong convexity and Lipschitz continuity hold for the whole $\mathbb{R}^N$; otherwise $e^x$ ($x\in\mathbb{R}$) is good enough in $[0,1]$.
-- New Remark: I ask this question because these two assumptions are often seen in optimization papers. Functions that satisfy one of the two are easy to think of; to satisfy these two assumptions at the same time, I really doubt how many functions exist. 
-- Finally came up with an example by myself:
$$f(x)=\log(x+\sqrt{1+x^2})+x^2,x\in\mathbb{R}.$$
Strong convexity: 
$$\nabla f(x) = \frac{1}{\sqrt{1+x^2}}+2x,$$
$$\nabla^2 f(x) = -\frac{x}{(1+x^2)^{3/2}}+2.$$
Since $$\left|\frac{x}{(1+x^2)^{3/2}}\right|=\left|\frac{x}{\sqrt{1+x^2}}\right|\frac{1}{1+x^2}<1,$$
strong convexity follows from $$3 > \nabla^2 f(x)>1.$$
Lipschitz continuous gradient:
\begin{align}
\left|\nabla f(x+h)-\nabla f(x)\right|&\stackrel{(a)}{=}\left|\nabla f(x)+\nabla^2 f(y)h-\nabla f(x)\right|\\
&=\left|\nabla^2 f(y)h\right|\\
&\leq 3|h|
\end{align}
where $(a)$ is from the mean value theorem and $y$ is some number in $[x,x+h]$ for $h\geq 0$ or $[x+h,x]$ for $h<0$.

Here's an example on $\mathbb R: f(x) = x^2-\cos x.$
A way to make lots of examples: Let $f$ be any positive bounded continuous function on $[0,\infty).$ For $x\ge 0,$ set
$$g(x) =\int_0^x \int_0^t f(s)\, ds\,dt.$$
Extend $g$ to an even function on all of $\mathbb R.$ Then $g$ satisfies the requirements. 
