I'm reading a proof about approximating lower semicontinuous functions by lipschitz functions. Commonly I read the formula:
$$u_t(x) = \inf_{y \in \mathbb{R}}\{u(y) + td(x,y) \}$$
I'm having trouble understanding what this function "looks like." For instance, if $u(y) = y^2$, if we want to compute $u_t(1)$, it appears that choosing $y=x=1$ will infimize the RHS. In fact, if I choose any $x \in \mathbb{R}$ then choosing $x = y$ will infimize RHS. So $u_t(x)$ just "tracks $u(y)$ with the vertex of an absolute value graph."
That doesn't seem useful for a proof for lower semicontinuity. Loosely speaking, how should I think of these functions $u_t(x)$?
In your example, your calculation is incorrect for certain values of $t$. If $t$ is large enough, then yes, you will have $y=x=1$, but when $t$ gets smaller you will need to choose a different $y$ to attain the infimum. When $u(x) = x^2$, and you want to calculate $u_t(x),$ note that $$u_t(x) = \inf_{y} \{ y^2 + t \lvert y-x\rvert\}.$$ You are minimizing $$f_t(y,x) = \left\{\begin{matrix} y^2 + ty - tx, & y < x, \\ y^2 - ty + tx, & y > x. \end{matrix}\right.$$ You need to check the points where $\partial_y f_t(y,1) = 0$ or the derivative is undefined; the potential points you need to check are, $y=t/2, y= -t/2$ or $y = x$ (though the first two points are only viable in certain cases).
Indeed, as you have it defined, you will find that $u_t(x)$ is Lipschitz with constant $L = t$. In general, if $u$ is smooth, you will find that $u_t(x) = u(x)$ at places where $\lvert u'(x)\rvert \le t$; otherwise, in regions where $\lvert u'(x) \rvert > t$, then $u_t(x)$ will essentially be linear with slope $t$, and it will be increasing or decreasing depending on the sign of $u'(x)$.
Going back to your example of $u(x) = x^2$, I have included four pictures to help you visualize this. The pictures are plotted for $x \in [-5,5]$. In each image, $u(x) = x^2$ is plotted in black and $u_t(x)$ is plotted in dotted red. Note that $\lvert u'(x)\rvert \le 10$ on this interval, thus when $t = 10$, we will have $u_t(x) \equiv u(x)$ on the interval. For lower values of $t$, $u_t(x) = u(x)$ for $x\in [-t/2,t/2]$ but outside that interval, $u_t(x)$ is essentially linear with slope of magnitude $t$.