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Given $f:X\to\mathbb{R}$ be a bounded function such that $\exists M$ s.t $|f(x)|\leq M$ and $f_k:X\to\mathbb{R}$ where $f_k=\inf_{y\in X}(f(y)+kd(x,y))$ Let $k\in\mathbb{N}$ and $X=\mathbb{R}$ and $d(x,y)$ be the usual metric. Let $(X,d)$ and $f$ be arbitary metric space and bounded function.

Show that if $f$ is both continuous and bounded then $f_k\to f$ pointwise as $k\to\infty$

How would I go on about proving this?

What I know so far is that.

Let $M>0$ using the contounity of $f$. Then for $x\in X$ and $\epsilon >0$, let $\delta >0$ and $f_k=\min{(T_1(x),T_2(x))}$, where

$$T_1(x) =\inf_{y\in B(x;\delta)}\{f(y)+kd(x,y)\},\qquad T_2(x) =T_1(x) =\inf_{y\notin B(x;\delta)}\{f(y)+kd(x,y)\} $$

How would I go on to prove that $T_1(x)\leq M$ and $T_2(x)\geq -M +k\delta$

Undergrad2019
  • 309

1 Answers1

1

First, $f_k$ is an increasing sequence as $f_k\leq f_{k+1}$ Also for $y=x$ we get that $f_k(x)=\underset{x\in X }{inf}(f(x)+kd(x,x))=\underset{x\in X}{inf}(f(x))\leq f(x)$

So $f_k$ is bounded as $f$ is so. This means that $f_k$ converges pointwisely.

Recall that in general if $inf A$ exists, then $\forall \epsilon >0$ $\exists x_0\in A$ such that $x_0\leq infA+\epsilon $

Applying this on $f_k(x)$, there exists a sequence $(y_k)$ of $X$ such that $f(y_k)+kd(x,y_k)\leq \frac{1}{k} + f_k(x)$ (took $\epsilon=\frac{1}{k}$)

Then $d(x,y_k)\leq \frac{f_k(x)+\frac{1}{k}-f(y_k)}{k}$. Notice that the right side converges to zero as both $f$ and $f_k$ are bounded, which means that $d(x,y_k)$ goes to zero that is $y_k$ converges to $x$.

So now that we already have $f(y_k)+kd(x,y_k)-\frac{1}{k} \leq f_k(x)\leq f(x)$ , take the limit on all sides.

We get $f(\lim_{k \to \infty }y_k)+0-0\leq \lim_{k \to \infty }f_k(x)\leq f(x)$.

(As $f$ is continuous, $\lim_{k \to \infty }f(y_k)=f(\lim_{k \to \infty }y_k)$)

Therefore we get, $f(x)\leq \lim_{k \to \infty }f_k(x)\leq f(x)$, hence the result.

Pedro Alvarès
  • 720