The Yosida transform and its properties. Let $\lambda \gt 0$ and $f:\Bbb R \to \Bbb R$. 
Define the Yosida transform of $f$ by
$$T_\lambda f(x) = \inf_{y \in \Bbb R}\{f(y) +\lambda |x-y|\}$$
So far I have showed that $T_\lambda f = \max\{g:g\le f$ and g is $\lambda$-Lipschitz$\}$.
Now, I want to show a few things about this operator: 


*

*if $f_n\to f$ pointwise on $\Bbb R$ then $T_\lambda f_n\to T_\lambda f$ pointwise on $\Bbb R$.

*if $f_n$ has a growth condition such as $f_n(x)\ge c|x|^p$ for some $p\gt 1$ and $f_n$ is convex for each $n$ then the convergence is uniform. 

*Suppose $f$ is convex and have the same growth as in (2). I'm wondering whether or not $T_\lambda f \to f$ as $\lambda \to \infty$.
I would really appreciate any help since I wasn't able to prove either claims.

My attempt: 
let $x\in \Bbb R$ we want to show that $\lim_n T_\lambda f_n(x) = T_\lambda f(x)$.
For each $n\in \Bbb N$ by the definition of the infimum, there is $(y_k^n)_{k=1}^{\infty}$ such that $\lim_{k \to \infty}(f(y_k^n) +\lambda |x-y_k^n| )= T_\lambda f_n(x)$.
I thought maybe looking at the "diagonal" $(y_n^n)_{n=1}^{\infty}$ but not sure if that helps since im not sure if this sequence converge.
Thanks for helping.
 A: Point 1 doesn't hold without additional assumptions. Consider
$$f_n(x) = \begin{cases} 0 &\text{if } x \neq \frac{1}{n}, \\ -1 &\text{if } x = \frac{1}{n}. \end{cases}$$
Then $f_n \to 0$ pointwise, but
$$T_{\lambda}f_n(x) = \begin{cases}\qquad 0 &\text{if } \bigl\lvert x - \frac{1}{n}\bigr\rvert \geqslant \lambda^{-1} \\ -1 + \lambda\bigl\lvert x - \frac{1}{n}\bigr\rvert &\text{if } \bigl\lvert x- \frac{1}{n}\bigr\rvert < \lambda^{-1}\end{cases}$$
converges pointwise to
$$h(x) = \begin{cases} \qquad 0 &\text{if } \lvert x\rvert \geqslant \lambda^{-1} \\ -1 + \lambda\lvert x\rvert &\text{if } \lvert x\rvert < \lambda^{-1} \end{cases}$$
and not to $0 = T_{\lambda}0$. We can modify the example to have the $f_n$ continuous (linearly interpolate between $0$ and $1/n$ and between $1/n$ and $2/n$, then $T_{\lambda}f_n$ is the same a above for $n > \lambda$), and if we multiply $f_n$ with a sequence of constants tending to $+\infty$, then $T_{\lambda}f_n \to -\infty$.
As we shall see below, one such assumption would be the convexity of all $f_n$.
We will need a few facts about convex functions. The fundamental fact is that for a convex $g \colon \mathbb{R} \to \mathbb{R}$, the difference quotient
$$Q(u,v) = \frac{g(v) - g(u)}{v-u}\,,$$
defined on $\{(u,v) \in \mathbb{R}^2 : u < v\}$ is (weakly) monotonically increasing in each variable separately, which can succinctly be stated as
$$r < s < t \implies Q(r,s) \leqslant Q(r,t) \leqslant Q(s,t)\,. \tag{1}$$
In fact, $(1)$ is equivalent to the convexity of $g$. Writing
$$s = \frac{t-s}{t-r}\cdot r + \frac{s-r}{t-r}\cdot t$$
the convexity of $g$ implies
$$g(s) \leqslant \frac{t-s}{t-r}g(r) + \frac{s-r}{t-r}g(t)\tag{2}$$
and hence
$$g(s) - g(r) \leqslant \biggl(\frac{t-s}{t-r}-1\biggr)g(r) + \frac{s-r}{t-r}g(t)
= \frac{r-s}{t-r}g(r) + \frac{s-r}{t-r}g(t) = \frac{s-r}{t-r}\bigl(g(t) - g(r)\bigr)\,.$$
Dividing by $s-r$ yields $Q(r,s) \leqslant Q(r,t)$. Similarly we can write the right hand side of $(2)$ as
$$g(t) - \frac{t-s}{t-r}\bigl(g(t) - g(r)\bigr)$$
and rearranging and dividing by $t-s$ yields $Q(r,t) \leqslant Q(s,t)$. Conversely, $r < s < t$ and $Q(r,s) \leqslant Q(s,t)$ yields
$$\frac{g(s)}{s-r} - \frac{g(r)}{s-r} \leqslant \frac{g(t)}{t-s} - \frac{g(s)}{t-s}
\iff g(s)\frac{(t-s) + (s-r)}{(t-s)(s-r)} \leqslant \frac{g(t)(s-r) + g(r)(t-s)}{(t-s)(s-r)}$$
and multiplication with $\frac{(t-s)(s-r)}{t-r}$ yields $(2)$. Since this holds for all $r < s < t$, $g$ is convex if $(1)$ holds.
Now $(1)$ implies that for every $x$ the right-hand derivative and left-hand derivative
$$D_{+}g(x) = \lim_{y \downarrow x} \frac{g(y) - g(x)}{y-x} = \inf_{y > x} \frac{g(y) - g(x)}{y-x} \quad\text{and}\quad D_{-}g(x) = \lim_{z \uparrow x} \frac{g(x) - g(z)}{x-z} = \sup_{z < x} \frac{g(x) - g(z)}{x-z}$$
exist, that $D_{-}g(x) \leqslant D_{+}g(x)$ holds for all $x$, and
$$x < y \implies D_{+}g(x) \leqslant D_{-}g(y)\,,$$
since $D_{+}g(x) \leqslant Q(x,y) \leqslant D_{-}g(y)$. In particular, $D_{+}g$ and $D_{-}g$ are (weakly) monotonically increasing. Further, $D_{+}g$ is right-continuous and $D_{-}g$ left-continuous, i.e. we have
$$D_{+}g(x) = \lim_{y\downarrow x} D_{+}g(y) = \inf_{y > x} D_{+}g(y)\quad \text{and}
\quad D_{-}g(x) = \lim_{z \uparrow x} D_{-}g(z) = \sup_{z < x} D_{-}g(z)\,.$$
For, given $\varepsilon > 0$ there is an $y > x$ such that $Q(x,y) < D_{+}g(x) + \varepsilon$. Then by continuity there is a $\xi \in (x,y)$ such that $Q(\xi,y) < D_{+}g(x) + 2\varepsilon$ and hence
$$D_{+}g(x) \leqslant D_{+}g(t) \leqslant Q(t,y) \leqslant Q(\xi,y) < D_{+}g(x) + 2\varepsilon$$
for all $t \in (x,\xi)$. The proof of left-continuity of $D_{-}g$ is similar. Thus
$$D_{+}g(x) = \inf_{y > x} D_{\pm} g(y) \quad \text{and}\quad D_{-}g(x) = \sup_{z < x} D_{\pm}g(z)\,.$$
$D_{-}g$ is continuous at $x$ if and only if $D_{+}g$ is continuous at $x$, if and only if $D_{-}g(x) = D_{+}g(x)$, if and only if $g$ is differentiable at $x$. This holds for all but countably many points. And $g$ is absolutely continuous,
$$g(y) - g(x) = \int_x^y g'(t)\,dt = \int_x^y D_{\pm}g(t)\,dt$$
for all $x,y$.
Now, given $\lambda > 0$, let
$$a = \sup \{x : D_{-}g(x) < -\lambda\} \quad \text{and} \quad b = \inf \{x : D_{+}g(x) > \lambda\}\,,$$
using the convention $\sup \varnothing = -\infty$, $\inf \varnothing = +\infty$ if needed. Then we have
$$T_{\lambda}g(x) = \begin{cases} g(a) + \lambda\lvert x-a\rvert &\text{if } x \leqslant a \\ \qquad g(x) &\text{if } a \leqslant x \leqslant b \\ g(b) + \lambda \lvert x-b\rvert &\text{if } b \leqslant x \end{cases} \tag{3}$$
if $a < +\infty$ and $b > -\infty$. If $a = +\infty$ or $b = -\infty$, then
$$T_{\lambda}g(x) = g(x) - \int_x^{+\infty} (-\lambda) - D_{\pm}g(t)\,dt \quad\text{or} \quad T_{\lambda}g(x) = g(x) - \int_{-\infty}^x D_{\pm}g(t) - \lambda\,dt$$
respectively. Typically these are both identically $-\infty$, but when $D_{\pm}g(x)$ approaches $\lambda$ fast enough as $x \to -\infty$ or $-\lambda$ fast enough as $x \to +\infty$ then $T_{\lambda}g$ is finite even if $D_{\pm}g(x) < -\lambda$ or $D_{\pm}g(x) > \lambda$ for all $x$. But we are interested in the case $a < +\infty$ and $b > -\infty$, i.e. there is an $x$ with $D_{-}g(x) \leqslant \lambda$ and a $y$ with $D_{+}g(y) \geqslant -\lambda$. Then the function on the right and side of $(3)$ is $\lambda$-Lipschitz, and it is $\leqslant g$. For $x > b$ we have
$$g(x) = g(b) + \int_b^x D_{\pm}g(t)\,dt > g(b) +\int_b^x \lambda\,dt = g(b) + \lambda \lvert x-b\rvert$$
and for $x < a$ we have
$$g(x) = g(a) - \int_x^a D_{\pm}g(t)\,dt > g(a) - \int_x^a -\lambda\,dt = g(a) +\lambda\lvert x-a\rvert\,.$$
Also, it is the largest $\lambda$-Lipschitz function $\leqslant g$.
For, suppose $h$ is such a function. Then clearly $h(x)$ is not larger than the right hand side of $(3)$ for $a \leqslant x \leqslant b$. And for $x > b$ we have
$$h(x) \leqslant h(b) + \lambda\lvert x-b\rvert \leqslant g(b) + \lambda\lvert x-b\rvert\,,$$
for $x < a$ we have
$$h(x) \leqslant h(a) + \lambda\lvert x-a\rvert \leqslant g(a) + \lambda\lvert x-a\rvert\,,$$
thus $h(x)$ doesn't exceed the right hand side of $(3)$ anywhere. By the characterisation of $T_{\lambda}g$ that you found, the right hand side of $(3)$ is indeed $T_{\lambda}g$.
Now it is convenient to step out of line and treat point 3 before 2. We now index $a$ and $b$ by $\lambda$. Then it is clear that $a_{\lambda} \leqslant -M < M \leqslant b_{\lambda}$ for $\lambda > \max \{ D_{+}g(M), -D_{-}g(-M)\}$ and consequently $T_{\lambda}g(x) = g(x)$ on $[-M,M]$. Thus
$$\lim_{\lambda \to +\infty} T_{\lambda}g = g$$
locally uniformly for convex $g$. Of course in general not uniformly on the whole of $\mathbb{R}$.
Also, we have $\lambda < \mu \implies T_{\lambda}f \leqslant T_{\mu}f$ (for general, not necessarily convex or even continuous $f$), hence
$$h(x) = \lim_{\lambda \to +\infty} T_{\lambda}f(x) = \sup_{\lambda > 0} T_{\lambda}f(x)$$
exists for every $x$, and $h$ is lower semicontinuous. So lower semicontinuity is a necessary condition for $T_{\lambda}f \to f$, but of course not sufficient. Neither is continuity sufficient, for $f(x) = -x^2$ we have $T_{\lambda}f \equiv -\infty$ for all $\lambda > 0$. We additionally need $T_{\lambda}f(x) > -\infty$ for sufficiently large $\lambda$. If that holds for one $x$, then it holds for all $x$, since $T_{\lambda}f$ is $\lambda$-Lipschitz. Thus we may look at $x = 0$. We have $T_{\lambda}f(0) \geqslant c$ if and only for all $x$
$$c \leqslant f(x) + \lambda\lvert x\rvert$$
holds. Then
$$\liminf_{\lvert x\rvert \to \infty} \frac{f(x)}{\lvert x\rvert} \geqslant \lim_{\lvert x\rvert \to \infty} \frac{c - \lambda\lvert x\rvert}{\lvert x\rvert} = -\lambda\,.$$
We thus have the additional necessary condition
$$\liminf_{\lvert x\rvert \to \infty} \frac{f(x)}{\lvert x\rvert} > -\infty\,. \tag{4}$$
If this holds and $f$ is lower semicontinuous, then $T_{\lambda}f \to f$ pointwise as $\lambda \to +\infty$. Since
$$\lim_{\lvert x\rvert \to \infty} \frac{\lvert x\rvert}{\lvert x-y\rvert} = 1$$
for every $y$, it suffices to prove that under the above assumptions $T_{\lambda}f(0) \to f(0)$. Let
$$\lambda_0 > -\liminf_{\lvert x\rvert \to \infty} \frac{f(x)}{\lvert x\rvert}$$
(and $\lambda_0 > 0$ of course). Then there is an $R > 0$ such that $f(x) > -\lambda_0\lvert x\rvert$ for $\lvert x\rvert \geqslant R$. Since $f$ is lower semicontinuous it attains its minimum, say $m$, on the compact set $[-R,R]$. Let $c < f(0)$. By lower semicontinuity there is a $\delta > 0$ such that $f(x) > c$ for $\lvert x\rvert < \delta$. Then $f(x) + \lambda\lvert x\rvert \geqslant f(x) > c$ for all $x$ with $\lvert x\rvert < \delta$ and all $\lambda > 0$. And
$$f(x) + \lambda\lvert x\rvert \geqslant m + \lambda\lvert x\rvert \geqslant m +\lambda \delta \geqslant c$$
for all $x$ with $\delta \leqslant \lvert x\rvert \leqslant R$ if $\lambda \geqslant \frac{c-m}{\delta}$. Finally, for $\lvert x\rvert \geqslant R$ and $\lambda > \lambda_0$ we have
$$f(x) + \lambda \lvert x\rvert = \bigl(f(x) + \lambda_0\lvert x\rvert\bigr) + (\lambda - \lambda_0)\lvert x\rvert > (\lambda - \lambda_0)\lvert x\rvert \geqslant (\lambda - \lambda_0)R \geqslant c$$
if $\lambda \geqslant \lambda_0 + \frac{c}{R}$. Thus $T_{\lambda}f(0) \geqslant c$ for all
$$\lambda > \max \biggl\{\lambda_0 + \frac{\lvert c\rvert}{R}, \frac{c-m}{\delta}\biggr\}\,.$$
Since $c < f(0)$ was arbitrary, $\lim T_{\lambda}f(0) \geqslant f(0)$. The other inequality is trivial since $T_{\lambda}f \leqslant f$.
Thus $T_{\lambda}f \to f$ pointwise if and only if $f$ is lower semicontinuous and satisfies $(4)$.
Now we come to point 2. First we note that the pointwise limit of a sequence of convex functions is convex. And in the situation of 2, $f$ also satisfies the growth condition
$$f(x) \geqslant c\lvert x\rvert^p\,. \tag{5}$$
Thus for $x > 0$ we have
$$D_{-}f(x) \geqslant Q(0,x) = \frac{f(x) - f(0)}{x} \geqslant \frac{c x^p - f(0)}{x} = c x^{p-1} - \frac{f(0)}{x}$$
and
$$D_{+}f(-x) \leqslant Q(-x,0) = \frac{f(0) - f(-x)}{x} \leqslant \frac{f(0) - c x^p}{x} = \frac{f(0)}{x} - cx^{p-1}\,,$$
hence $D_{\pm}f(x) > \lambda$ and $D_{\pm}f(-x) < -\lambda$ for all sufficiently large $x$. Therefore $-\infty < a \leqslant b < +\infty$ for $a$ and $b$ defined as above. This is the only way $(5)$ is used, thus we might replace $(5)$ with the weaker
$$\lim_{x \to \infty} D_{+}f(x) > \lambda \quad\text{and}\quad \lim_{x \to -\infty} D_{-}f(x) < -\lambda\,. \tag{6}$$
Now we need an additional fact about convex functions. If $(f_n)$ is a sequence of convex functions converging pointwise to $f$, then
\begin{align}
\limsup_{n \to \infty} D_{+}f_n(x) &\leqslant D_{+}f(x)\,, \\
\liminf_{n \to \infty} D_{-}f_n(x) &\geqslant D_{-}f(x)\,.
\end{align}
Choose $h > 0$. Then
$$D_{+}f_n(x) \leqslant \frac{f_n(x+h) - f_n(x)}{h}$$
for every $n$, and hence
$$\limsup_{n \to \infty} D_{+}f_n(x) \leqslant \lim_{n \to \infty} \frac{f_n(x+h) - f_n(x)}{h} = \frac{f(x+h) - f(x)}{h}$$
holds for every $h > 0$. Consequently
$$\limsup_{n \to \infty} D_{+}f_n(x) \leqslant \inf_{h > 0} \frac{f(x+h) - f(x)}{h} = D_{+}f(x)\,.$$
The proof of the second inequality is analogous.
In particular, at all points $x$ where the limit $f$ is differentiable, we have
$$\lim_{n \to \infty} D_{-}f_n(x) = \lim_{n\to \infty} D_{+}f_n(x) = f'(x)$$
regardless of whether any $f_n$ is differentiable there.
Now consider an arbitrary compact interval $[u,v]$. By making $u$ slightly smaller and $v$ slightly larger, if necessary, we can assume that $f$ is differentiable at $u$ and at $v$. Thus $D_{\pm}f_n(w) \to f'(w)$ for $w = u$ and $w = v$. Let $M = \max \{ \lvert f'(u)\rvert, \lvert f'(v)\rvert\}$. Then there is an $n_0$ such that $\lvert D_{\pm}f_n(w)\rvert \leqslant M+1$ for all $n \geqslant n_0$, where $w$ again is either of $u$ and $v$. By monotonicity of the one-sided derivatives of convex functions,
$$-M-1 \leqslant D_{\pm}f_n(x) \leqslant M+1$$
for all $n \geqslant n_0$ and all $x \in [u,v]$. Thus the family $\{ f_n : n \geqslant n_0\} \cup \{f\}$ is equilipschitz, a fortiori uniformly equicontinuous, on $[u,v]$. (The restriction $n \geqslant n_0$ is unnecessary, the same conclusion holds for the entire sequence, just with a possibly larger Lipschitz constant.) Therefore $f_n \to f$ uniformly on $[u,v]$.
Back to our situation with $(6)$. In addition to $a$ and $b$, define
$$\alpha = \max \{x : D_{-}f(x) \leqslant -\lambda\} \quad \text{and}\quad \beta = \min \{x : D_{+}f(x) \geqslant \lambda\}\,.$$
Because of the left- and right-continuity of $D_{-}f$ and $D_{+}f$ respectively one can use $\max$ and $\min$ here. Then we have
$$ -\infty < a \leqslant \alpha \leqslant \beta \leqslant b < +\infty\,.$$
Each of the non-strict inequalities here can actually be an equality, but they all can also be strict.
We shall first prove the uniform convergence of $T_{\lambda}f_n$ to $T_{\lambda}f$ on $[b,+\infty)$, noting that the uniform convergence on $(-\infty,a]$ follows in the same way. Then we prove uniform convergence on $[\beta,b]$, again with uniform convergence on $[a,\alpha]$ following by the same argument. Finally we prove uniform convergence on $[\alpha,\beta]$. Since there are only finitely many parts, uniform convergence on $\mathbb{R}$ follows.
Let $\varepsilon > 0$ be given. First note that in the formula $(3)$ for $T_{\lambda}f$ we can replace $a$ with $\alpha$ and $b$ with $\beta$ because $f'(x) = -\lambda$ for $a < x < \alpha$ and $f'(x) = \lambda$ for $\beta < x < b$. Thus for $x\geqslant \beta$ we have
$$T_{\lambda}f_n(x) \leqslant f_n(\beta) + \lambda (x-\beta) = T_{\lambda}f(x) + \bigl(f_n(\beta) - f(\beta)\bigr) \leqslant T_{\lambda}f(x) + \lvert f_n(\beta) - f(\beta)\rvert $$
and that is $\leqslant T_{\lambda}f(x) + \varepsilon$ for $n \geqslant n_1$.
Now choose $b < c < b + \frac{\varepsilon}{3\lambda}$ such that $f$ is differentiable at $c$. Since $f'(c) > \lambda$ by definition of $b$, we have $D_{-}f_n(c) > \lambda$ for $n \geqslant n_2$. Also, pick $d <\beta$ such that $f$ is differentiable at $d$. Then $f'(d) < \lambda$ by definition of $\beta$, and $D_{+}f_n(d) < \lambda$ for $n \geqslant n_3$. It follows that $d \leqslant b_n \leqslant c$ for $n \geqslant n_4 = \max \{n_2, n_3\}$, where $b_n$ is defined analogously to $b$. For $x\geqslant c$ and $n \geqslant n_4$ we then have
$$T_{\lambda}f_n(x) = f_n(b_n) + \lambda(x - b_n) = f(b_n) + \lambda(x-b_n) + \bigl(f_n(b_n) - f(b_n)\bigr) \geqslant T_{\lambda}f(x) - \lvert f_n(b_n) - f(b_n)\rvert\,.$$
Since $f_n \to f$ uniformly on $[d,c]$, the right hand side is $\geqslant T_{\lambda}f(x) - \varepsilon/3$ for $n \geqslant n_5$. For $b \leqslant x\leqslant c$ we have
$$T_{\lambda}f(x) - T_{\lambda}f_n(x) \leqslant T_{\lambda}f(c) - T_{\lambda}f_n(c) + 2\lambda\lvert x-c\rvert  \leqslant \frac{\varepsilon}{3} + 2\lambda\frac{\varepsilon}{3\lambda} = \varepsilon\,.$$
Thus $\lvert T_{\lambda}f(x) - T_{\lambda}f_n(x)\rvert \leqslant \varepsilon$ on $[b,+\infty)$ for $n \geqslant \max \{n_1,n_5\}$.
Next, for $\beta \leqslant x \leqslant b$ and $n \geqslant \max n_5$ we have
$$T_{\lambda}f_n(x) \geqslant T_{\lambda}f_n(b) - \lambda(b-x) \geqslant T_{\lambda}f(b) - \varepsilon - \lambda(b-x) = T_{\lambda}f(x) - \varepsilon$$
since $T_{\lambda}f_n$ is $\lambda$-Lipschitz and $T_{\lambda}f(x) = f(\beta) + \lambda(x-\beta)$ on $[\beta,b]$. Therefore $\lvert T_{\lambda}f_n(x) - T_{\lambda}f(x)\rvert \leqslant \varepsilon$ on $[\beta,b]$ for $n \geqslant \max \{n_1,n_5\}$. In the same way it follows that there is an $n_6$ such that $\lvert T_{\lambda}f_n(x) - T_{\lambda}f(x)\rvert \leqslant \varepsilon$ on $(-\infty,\alpha]$ for $n \geqslant n_6$.
Finally, assuming $\alpha < \beta$ since for $\alpha = \beta$ there is nothing to prove, pick $\alpha < c < \alpha + \frac{\varepsilon}{3\lambda}$ and $\beta - \frac{\varepsilon}{3\lambda} < d < \beta$ such that $c < d$ and $f$ is differentiable at $c$ and at $d$. Since $-\lambda < f'(c) \leqslant f'(d) < \lambda$, we have $D_{-}f_n(c) > -\lambda$ and $D_{+}f_n(d) < \lambda$ for $n \geqslant n_7$. Thus 
$T_{\lambda}f_n = f_n$ on $[c,d]$ for $n \geqslant n_7$, and
$$\lvert T_{\lambda}f_n(x) - T_{\lambda}f(x)\rvert = \lvert f_n(x) - f(x)\rvert \leqslant \frac{\varepsilon}{3}$$
on $[c,d]$ for $n \geqslant n_8$. For $d \leqslant x \leqslant \beta$ we have
$$\lvert T_{\lambda}f_n(x) - T_{\lambda}f(x)\rvert \leqslant \lvert T_{\lambda}f_n(d) - T_{\lambda}f(d)\rvert + 2\lambda\lvert x-d\rvert \leqslant \frac{\varepsilon}{3} + 2\lambda\frac{\varepsilon}{3\lambda} = \varepsilon\,.$$
Similarly for $\alpha \leqslant x \leqslant c$, so $\lvert T_{\lambda}f_n(x) - T_{\lambda}f(x)\rvert \leqslant \varepsilon$ on $[\alpha,\beta]$ for $n \geqslant n_8$.
Altogether $\lvert T_{\lambda}f_n(x) - T_{\lambda}f(x)\rvert \leqslant \varepsilon$ on all of $\mathbb{R}$ for $n \geqslant \max \{n_1, n_5, n_6, n_8\}$.
