How to solve $u(x,t) = \min_{y\in \mathbb R^n} \left\{ \frac{|x-y|^2}{2t}+ |y|\right\}$?

Question

For $x\in \mathbb R^n ,t >0 $, let $$ u(x,t) = \min_{y\in \mathbb R^n} \left\{ \frac{|x-y|^2}{2t}+ |y|\right\}. \tag{1} $$ Then, How to show that $$ u(x,t)=\left\{ \begin{aligned} &|x|-t/2 & \quad |x|\ge t\\ &\frac{|x|^2}{2t} & \quad |x|\le t \end{aligned} \right. \tag{2} $$

I get the problem from 134th page of Evans' Partial Differential Equations. The author say that, when $|x|>t$, there is $$ D_y \left(\frac{|x-y|^2}{2t}+ |y|\right) = \frac{y-x}{t} + \frac{y}{|y|} \tag{3} $$ I have no idea at all why we need $|x|>t$ to get (3). In fact, $|x-y|^2 = \sum_{i=1}^n (x_i- y_i)^2$, therefore $$ D_y |x-y|^2 =2(y-x) \tag{4} $$ Similarly, there is $D_y|y| = \frac{y}{|y|}$. So, I can't understand why we need $|x|>t$ ?

On the other hand, the author say, when $|x| \le t$, the minimum is attained at $y=0$. I have no clue either.

Besides, there is a similar problem, but its answer doesn't solve my confusion.

Answer 1

Fix $x\in \mathbb R^n$ and $t>0$, and let $$v(y) = \frac{\vert x- y\vert^2}{2t}+\vert y \vert \qquad \text{for all }y\in \mathbb R^n. $$

Firstly, observe that $$v(y)\geq \frac{(\vert y \vert - \vert x \vert)^2}{2t} +\vert y\vert \to +\infty \text{ as }\vert y \vert \to +\infty,$$ so $v$ must attain a global minimum in some ball $B_R$.

If $y \neq 0$ then \begin{align*} \nabla_yv(y)&= \frac{y-x}{t} + \frac{y}{\vert y \vert }= \bigg ( \frac 1 t + \frac 1 {\vert y \vert}\bigg)y- \frac x t. \end{align*} If $y$ is a critical point then $\nabla_yv(y)=0$, so $$\bigg ( \frac 1 t + \frac 1 {\vert y \vert}\bigg)y= \frac x t. \tag{$\ast$}$$ Taking the norm of both sides then doing some simple rearranging, we obtain $$ \vert y \vert =\vert x \vert-t . $$

Thus, if $\vert x \vert \leq t$ then there are no critical points in $B_R\setminus \{0\}$, so the minimum must be attained at $y=0$ and so $$u(x,t) = \frac{\vert x \vert^2}{2t}. $$

Now, if $\vert x \vert > t$ then the minimum could still occur when $y=0$, or at some point satisfying $\vert y \vert = \vert x \vert-t$. Returning to ($\ast$), we have \begin{align*} y&=\bigg ( \frac 1 t + \frac 1 {\vert y \vert}\bigg)^{-1} \frac x t \\ &= \bigg ( \frac 1 t + \frac 1 {\vert x \vert -t}\bigg)^{-1} \frac x t \\ &= \bigg( 1-\frac{t }{\vert x \vert}\bigg)x. \end{align*} In this case, $$v(y) = \frac t2+ \vert x \vert -t = \vert x \vert - \frac t 2. $$ Since, when $\vert x\vert>t$, $$ \frac{\vert x \vert^2}{2t}> \vert x \vert - \frac t 2,$$ we have in this case $$ u(x,t) = \vert x \vert - \frac t 2$$ as required.

Answer 2

function $f(y)$ is made of two pieces: $\frac{|x-y|^2}{2t}$ and $|y|$. the first piece is like a quadratic bowl centered at $y=x$. it's smooth and strictly convex since $t>0$. the second piece, $|y|$, is the standard norm. it's convex, but it has a sharp point at $y=0$, so it's not differentiable there. when you add a strictly convex function and a convex function, you get a strictly convex function. this is great because it means $f(y)$ has a unique minimum point, let's call it $y^*$. also, $f(y)$ grows large as $|y|$ gets large because of the $|y|^2$ term hidden in $|x-y|^2$, so a minimum must exist.

here's how we find this minimum $y^*$. the standard approach is to find where the gradient is zero. let's compute the gradient of $f(y)$ with respect to $y$: $\nabla_y \left( \frac{|x-y|^2}{2t} \right) = \frac{1}{2t} \nabla_y((x-y)\cdot(x-y)) = \frac{1}{2t} (2(x-y)\cdot(-1)) = \frac{y-x}{t}$. the gradient of $|y|$ is $\nabla_y |y| = \frac{y}{|y|}$. however, the gradient of $|y|$ only exists if $y \neq 0$.

so, let's consider two possibilities for the minimum $y^*$:

case 1: the minimum occurs at $y^* \neq 0$. if the minimum isn't at the origin, then $f(y)$ is differentiable at $y^*$. in this case, the minimum must be where the gradient is zero: $$ \nabla_y f(y^*) = \frac{y^*-x}{t} + \frac{y^*}{|y^*|} = 0 \quad (*) $$ this is exactly your equation (3). so, this equation relies on the assumption that the minimum $y^*$ is not zero.

solve (*) for $y^*$. $$ \frac{y^*}{t} + \frac{y^*}{|y^*|} = \frac{x}{t} $$ $$ y^* \left( \frac{1}{t} + \frac{1}{|y^*|} \right) = \frac{x}{t} $$ $$ y^* = \frac{x/t}{1/t + 1/|y^*|} = \frac{x}{1 + t/|y^*|} $$ this tells us that $y^*$ must be a positive multiple of $x$. let's write $y^* = c x$ for some $c > 0$. (we need $x \neq 0$ for $y^* \neq 0$. if $x=0$, the equation becomes $\frac{y^*}{t} + \frac{y^*}{|y^*|} = 0$, which has no solution for $y^* \neq 0$ since $t>0$ and $|y^*|>0$). substituting $y^* = c x$ into the equation (assuming $x \neq 0$): $$ c x = \frac{x}{1 + t/|c x|} = \frac{x}{1 + t/(c|x|)} $$ divide by $x$ (since $x \neq 0$): $$ c = \frac{1}{1 + t/(c|x|)} $$ $$ c (1 + t/(c|x|)) = 1 $$ $$ c + \frac{t}{|x|} = 1 $$ $$ c = 1 - \frac{t}{|x|} $$ so, the potential minimizer is $y^* = \left(1 - \frac{t}{|x|}\right) x$.

now, for our assumption $y^* \neq 0$ to be valid we need $c > 0$. $c > 0 \implies 1 - \frac{t}{|x|} > 0 \implies 1 > \frac{t}{|x|} \implies |x| > t$. so, the gradient equation (*) only makes sense and gives us a non-zero solution $y^*$ when $|x| > t$. this is why evans mentions $|x| > t$ in relation to equation (3) – it's the condition under which the minimum occurs away from the non-differentiable point $y=0$.

if $|x| > t$, let's calculate the minimum value $u(x,t) = f(y^*)$: we have $y^* = (1 - t/|x|) x$. $x - y^* = x - (1 - t/|x|) x = (t/|x|) x$. $|x - y^*| = |(t/|x|) x| = (t/|x|) |x| = t$. $|y^*| = |(1 - t/|x|) x| = |1 - t/|x|| |x|$. since $|x| > t$, we know $1 - t/|x|$ is positive, so $|y^*| = (1 - t/|x|) |x| = |x| - t$. plugging these in: $$ u(x, t) = \frac{|x-y^*|^2}{2t} + |y^*| = \frac{t^2}{2t} + (|x| - t) = \frac{t}{2} + |x| - t = |x| - \frac{t}{2} $$ this matches the first case in formula (2).

case 2: the minimum occurs at $y^* = 0$. what if $|x| \le t$? our calculation above didn't yield a valid $y^* \neq 0$. maybe the minimum is at $y=0$? we can't use the gradient being zero here, because $f(y)$ isn't differentiable at $y=0$. however, for a convex function like $f(y)$, the minimum occurs at $y^*$ if and only if the zero vector is in the 'subdifferential' of $f$ at $y^*$. the subdifferential $\partial f(y^*)$ is like a generalization of the gradient for non-smooth points. the subdifferential of $f(y) = g(y) + h(y)$ where $g(y)=\frac{|x-y|^2}{2t}$ and $h(y)=|y|$ is $\partial f(y) = \nabla g(y) + \partial h(y)$. at $y=0$, $\nabla g(0) = \frac{0-x}{t} = -x/t$. the subdifferential of $h(y)=|y|$ at $y=0$ is the set of all vectors $p$ such that $|p| \le 1$. think of it as all possible 'slopes' at the sharp point of the cone $|y|$. let's call this set $B_1(0)$. so, $\partial f(0) = -x/t + B_1(0)$. the minimum is at $y^*=0$ if $0 \in \partial f(0)$. $$ 0 \in -\frac{x}{t} + \{ p \in \mathbb{R}^n : |p| \le 1 \} $$ this means there exists a $p$ with $|p| \le 1$ such that $0 = -x/t + p$. this is the same as saying $p = x/t$ for some $p$ with $|p| \le 1$. the condition is simply $|x/t| \le 1$, which means $|x| \le t$.

so, if $|x| \le t$, the minimum of $f(y)$ occurs at $y^* = 0$. let's find the minimum value in this case: $$ u(x, t) = f(0) = \frac{|x-0|^2}{2t} + |0| = \frac{|x|^2}{2t} $$ this matches the second case in formula (2).

so we found that:

• if $|x| > t$, the minimum is $u(x, t) = |x| - t/2$ (attained at $y^* = (1-t/|x|)x$).
• if $|x| \le t$, the minimum is $u(x, t) = |x|^2 / (2t)$ (attained at $y^* = 0$).

this is exactly the formula (2): $$ u(x, t) = \begin{cases} |x| - t/2 & \text{if } |x| \ge t \\ \frac{|x|^2}{2t} & \text{if } |x| \le t \end{cases} $$ (note that the formulas agree when $|x|=t$, both giving $t/2$).

so, the reason $|x|>t$ was needed for equation (3) is that's the condition for the minimum to happen away from $y=0$, where you can use the regular gradient. when $|x| \le t$, the pull towards $y=x$ from the first term isn't strong enough to overcome the preference of the $|y|$ term for $y=0$, so the minimum stays at the origin.

Answer 3

Clearly, $u(0,t)=0$. Assume now $x\ne0$. By Pythagoras' theorem, the quantity depending on $y$ (for $x$ and $t$ fixed) is smaller when $y=kx$ for some $k\ge0$, so $$u(x,t)=\min_{k\ge0}\left(\frac{|x-kx|^2}{2t}+k|x|\right)=\min_{k\ge0}\left(\frac{|x|^2}{2t}(k-k_0)^2+|x|-\frac t2\right),$$where$$k_0=1-\frac t{|x|}.$$Therefore:

• if $k_0\ge0$, $u(x,t)=|x|-\frac t2$;
• if $k_0\le0$, $u(x,t)=\frac{|x-0x|^2}{2t}+0|x|=\frac{|x|^2}{2t}$.

Answer 4

Let $a=|x|,b=|y|$ and $\theta$ be angles between $x$ and $y$, then $$|x-y|^2=a^2+b^2-2ab\cos\theta\ge a^2+b^2-2ab$$ $$\implies \frac{|x-y|^2}{2t}+|y|\ge\frac{a^2+b^2-2ab}{2t}+b=\frac{b^2+2b(t-a)+(t-a)^2-t^2+2at}{2t}\\ \ge\frac{-t^2+2at}{2t}=a-\frac{t}{2}$$ Equality happens when $\theta=0$ and $b=a-t$ or $|y|=|x|-t$.

Case 1: When $|x|\ge t$ equality do happens.

Case 2: When $|x|<t$ equality do not happens, then the minimum of the quadratic in $b$ with positive leading coefficient $$\frac{a^2+b^2-2ab}{2t}+b$$ Occurs at the boundary of $b\ge 0$ or $b=0$ or $y=0$.