Uniform convergence and differentiability

[gugo82]

Let:

$\chi (x):=\{(1, " if " |x|<1), (0, " otherwise" ):}$

and for each $n\in \NN$ define:

$\chi_n (x):=n/2 \chi (nx)$

so that $\chi_n \geq 0$, $"supt" \chi_n =[-1/n,1/n]$* and $\int_RR \chi_n (x)" d"x =1$.
Choose $f: RR \to RR$ bounded and Riemann integrable on every bounded interval $[a,b]\subset RR$ and put:

(*) $f_n(x):= \int_RR \chi_n(x-y) f(y) " d"y\quad$.

Prove that:

1. (*) defines a bounded Lipschitz continuous function for each $n$;

2. if $f\in C(RR)$, then each $f_n$ is of class $C^1$;

3. if $f\in C^1(RR)$, then each $f_n$ is of class $C^2$;

4. if $f\in C(RR)$, then $f_n \to f$ uniformly on compacts;

5. if $f$ is uniformly continuous in $RR$, then $f_n\to f$ uniformly in $RR$;

6. if $f \in C^1(RR)$, then $f'_n \to f'$ uniformly on bounded sets.

__________
* Here $"supt" \chi_n :=\bar(\{ x\in RR: \quad \chi_n(x)!=0\})$ is the support of $\chi_n$.

[gugo82]

"Gugo82":6. if $f \in C^1(RR)$, then $f'_n \to f'$ uniformly on bounded sets.

By (4) we have:

$\quad f'_n(x)=n/2[f(x+1/n)-f(x-1/n)]=1/2*[(f(x+1/n)-f(x))/(1/n)+(f(x-1/n)-f(x))/(-1/n)]$

so that:

$\quad \lim_n f'_n(x)=1/2*[f'(x)+f'(x)]=f'(x)$

and $f'_n\to f'$ pointwise in $RR$.
Now we have to show that the convergence is uniform on bounded sets: let $B\subset RR$ be bounded, denote with $\barB$ it's closure (which is compact), put $B_n:=\{y\in RR: " dist"(y,B)<=1/n\}$ and note that $\barB_n$ is compact and that $B\subseteq \barB \subset \barB_(n+1)\subseteq \barB_n \subseteq \barB_1$; for each $x\in B$ we have also $[x-1/n,x+1/n]\subseteq \barB_n\subseteq \barB_1$.
By Lagrange's Theorem we can find $\xi_n \in [x-1/n,x+1/n]\subseteq \barB_n$ s.t.:

(10) $\quad |f'_n(x)-f'(x)|=|f'(\xi_n)-f'(x)|$

and by Cantor's Theorem $f'$ is uniformly continuous on $\barB_1$: therefore we can find $\delta_\epsilon >0$ in such a way that for all $x\in B$ and for all $y$ with $|x-y|<\delta_\epsilon$ it comes $|f'(y)-f'(x)|<\epsilon/2$ ($\epsilon >0$). If we choose $n>\nu_\epsilon:=[1/\delta_\epsilon]+1$, we have $[x-1/n,x+1/n]\subseteq [x-\delta_\epsilon,x+\delta_\epsilon]$ and $|\xi_n-x|<\delta_\epsilon$ so that from (10) we get:

(11) $\quad |f'_n(x)-f'(x)|<\epsilon/2 \quad$;

taking the l.u.b.s over $B$ on both sides of (11), we find that:

$\quad "sup"_B |f'_n-f'|<\epsilon$

for $n>\nu_\epsilon$ and finally $f'_n\to f'$ uniformly on $B$, which was our claim.

[gugo82]

"Gugo82":4. if $f\in C(RR)$, then $f_n \to f$ uniformly on compacts;

Let $K\subset RR$ be a fixed compact set, let $K_n:=\{y\in RR: "dist"(y,K)<=1/n\}$ (here $"dist"(y,K):="inf"_(x\in K)|x-y|$) and note that $K_n$ is itself compact and that $K\subset K_(n+1)\subset K_n\subseteq K_1$ for all $n\in NN$.
We choose $x \in K$ and write:

$\quad |f_n(x)-f(x)|=|\int_RR \chi_n(x-y) f(y)" d"y-f(x)*\int_RR\chi_n(x-y)" d"y|<=\int_RR\chi_n(x-y)*|f(y)-f(x)|" d"y \quad$;

since $x \in K$, we got $"supt"\chi_n (x-y) \subseteq K_n \subseteq K_1$, therefore we can write:

(7) $\quad |f_n(x)-f(x)|<=\int_(K_1) \chi_n(x-y)*|f(y)-f(x)|" d"y$

with the compact $K_1$ which do not depend on $x$. By Cantor's Theorem, $f$ is uniformly continuous on $K_1$, meaning that:

$\quad AA \epsilon>0 ,\exists \delta_\epsilon>0 :\quad AAx,y \in K_1 " with " |x-y|<\delta_\epsilon , |f(y)-f(x)|<\epsilon/2 \quad$;

now fix $\epsilon >0$: we have $|x-y|<\delta_\epsilon$ for all $x\in K$ and $y\in K_n$ if the index $n$ is chosen in such a way that $1/n<\delta_\epsilon$, that is to say $n>\nu_\epsilon:=[1/\delta_\epsilon]+1$. Then for $n>\nu_\epsilon$ we have:

(8) $\quad |f_n(x)-f(x)|<\epsilon/2 \int_(K_1) \chi_n(x-y)" d"y=\epsilon/2$

for all $x\in K$. Taking the least upper bounds over $K$ of the first and last member of (8) we find:

$"sup"_K |f_n-f|<=\epsilon/2<\epsilon$

for $n>\nu_\epsilon$, which means $f_n\to f$ uniformly on $K$.

"Gugo82":5. if $f$ is uniformly continuous in $RR$, then $f_n\to f$ uniformly in $RR$;

As above we have:

$\quad |f_n(x)-f(x)|<=n/2\int_(x-1/n)^(x+1/n)|f(y)-f(x)|" d"y$

for every $x\in RR$; by uniform continuity, chosen $\epsilon>0$ we can find $\delta_\epsilon>0$ s.t. $|f(y)-f(x)|<\epsilon/2$ when $|x-y|<\delta_\epsilon$: if we select $n>\nu_\epsilon:=[1/\delta_\epsilon]+1$, then $[x-1/n,x+1/n] \subseteq [x-\delta_\epsilon,x+\delta_\epsilon]$ and:

(9) $\quad |f_n(x)-f(x)|<=n/2\epsilon/2*\int_(x-1/n)^(x+1/n)" d"y=\epsilon/2 \quad$.

Taking the l.u.b.s over $RR$ of the first and last member of (9) we find:

$\quad "sup"_RR |f_n-f| <\epsilon$

for all $n>\nu_\epsilon$ so that $f_n\to f$ uniformly in $RR$.

[gugo82]

"Gugo82":2. if $f\in C(RR)$, then each $f_n$ is of class $C^1$;

From (*) and the very definition of $\chi_n$ we got:

$f_n(x)=n/2\int_(x-1/n)^(x+1/n)f(y)" d"y \quad$;

if $f\in C(RR)$, the Fundamental Theorem of Integral Calculus says that $f_n$ is differentiable everywhere in $RR$ and that:

(4) $\quad f'_n(x)=n/2*[f(x+1/n)-f(x-1/n)]$

so that $f_n \in C^1(RR)$.

"Gugo82":3. if $f\in C^1(RR)$, then each $f_n$ is of class $C^2$;

If $f\in C^1(RR)$, we can differentiate both sides of (4) to get:

(5) $\quad f''_n(x)=n/2*[f'(x+1/n)-f'(x-1/n)] \quad$,

so that $f_n \in C^2(RR)$.
Furthermore, if we suppose $f\in C^k(RR)$ for $k\in NN$, differentiating both sides of (4) $k$ times yelds $f_n\in C^(k+1)(RR)$ and:

(6) $f_n^((k+1))(x)=n/2*[f^((k))(x+1/n)-f^((k))(x-1/n)] \quad$.

[gugo82]

I'd like to propose my solutions; maybe someone can check them.

"Gugo82":Choose $f: RR \to RR$ bounded and Riemann integrable on every bounded interval $[a,b]\subset RR$ and put:

(*) $f_n(x):= \int_RR \chi_n(x-y) f(y) " d"y\quad$.

Prove that:

1. (*) defines a bounded Lipschitz continuous function for each $n$;

As elgiovo wrote, each $f_n$ is bounded. Infact:

$|f_n(x)|=n/2|\int_(-1/n)^(1/n)\chi(y) f(x-y)" d"y|<= n/2*||f||_oo\int_(-1/n)^(1/n)" d"y=||f||_oo$

holds for every $x\in RR$; so $||f_n||_oo<=||f||_oo$.

Now we turn to Lipschitz continuity: let $x_1,x_2 \in RR$ and we have:

(1) $|f_n(x_1)-f_n(x_2)|<=\int_RR |\chi_n(x_1-y)-\chi_n(x_2-y)|*|f(y)|" d"y<=||f||_oo*\int_RR |\chi_n(x_1-y)-\chi_n(x_2-y)|" d"y \quad$.

The supports of $\chi_n(x_1-y)$ and $\chi_n(x_2-y)$ are, respectively, $[x_1-1/n,x_1+1/n]$ and $[x_2-1/n,x_2+1/n]$: these two sets overlap (i.e. they have a common inner point) iff $|x_1-x_2|<2/n$, so we can distinguish between two cases:

a) $|x_1-x_2|>=2/n$: in this case the supports of $\chi_n(x_1-y)$ and $\chi_n(x_2-y)$ do not overlap and we got:
$\quad |\chi_n(x_1-y)-\chi_n(x_2-y)|=\{(n/2, " if " x_1-1/n and a direct calculation yelds:
(2) $\quad \int_RR |\chi_n(x_1-y)-\chi_n(x_2-y)|" d"y=n/2*(2/n+2/n)=2<=n|x_1-x_2| \quad$;

b) $|x_1-x_2|<2/n$: in this case the supports of $\chi_n(x_1-y)$ and $\chi_n(x_2-y)$ do overlap; if $x_1<=x_2$, we got:
$\quad |\chi_n(x_1-y)-\chi_n(x_2-y)|=\{(n/2, " if " x_1-1/n so we can evaluate:
$\quad \int_RR |\chi_n(x_1-y)-\chi_n(x_2-y)|" d"y=n/2*(x_2-x_1+x_2-x_1)=n(x_2-x_1) \quad$;
otherwise, if $x_2<=x_1$, by reasoning as above we have:
$\quad \int_RR |\chi_n(x_1-y)-\chi_n(x_2-y)|" d"y=n(x_1-x_2) \quad$;
putting these two equalities together yelds (2) again.

It follows from a-b) that the estimate (2) is valid for every choice of $x_1,x_2 \in RR$, so we can use it to bound by above the last member of (1). Actually it's:

(3) $\quad |f_n(x_1)-f_n(x_2)|<=||f||_oo*\int_RR |\chi_n(x_1-y)-\chi_n(x_2-y)|" d"y<=n||f||_oo*|x_1-x_2|$

so that $f_n$ is (globally) Lipschitz with constant $L_n=n||f||_oo$.

[gugo82]

Boundness: since $|f_n(x)|<= n/2||f||_oo \int_("supt" \chi_n) " d"t=||f||_oo$ for all $x\in RR$, each $f_n$ is bounded.

Lipschitz continuity: a function $F$ is said to be locally Lipschitz continuous iff for each $x_0\in RR$ exist $\delta>0$ and $C_(x_0) >=0$ s.t. $|F(x_1)-F(x_2)|<=C_(x_0) |x_1-x_2|$ for all $x_1,x_2 \in ]x_0-\delta,x_0+\delta[$; a function is said to be (globally) Lipschitz continuous iff exists $C>=0$ (which doesn't depend on points of $RR$) s.t. $|F(x_1)-F(x_2)|<=C |x_1-x_2|$ for all $x_1,x_2 \in RR$.
The problem asks to prove that $f_n$ is globally Lipschitz; your reasoning leads to the inequality:

$|f_n(x_1)-f_n(x_2)|<= n/2\int_(-1/n)^(1/n) |f(x_1-y)-f(x_2-y)|" d"y$

so you have to obtain $"r.h.s."<=C|x_1-x_2|$ (with $C>=0$ which doesn't depend on $x_1,x_2$) in order to conclude that $f_n$ is globally Lipschitz.
Your conclusion is wrong since the inequality $2||f||_oo<=|x_1-x_2|$ fails in defining a neighbourhood of $x_1$ or $x_2$ (for local Lipschitz continuity) or a constant independent on points of $RR$ (for global Lipschitz continuity).

[elgiovo]

"Gugo82":
Do you mean that the last member has a least upper bound in absolute value?

Yes, that is what I wanted to say. Is it wrong?

"Gugo82":
Sorry, I can't understand. Do you mean that a bounded function is always locally Lipschitz?

For each $x_1,x_2$ must be $2|| f ||_(oo)<=|x_1-x_2|$, so the (local) Lipschitz constant is $(2|| f ||_(oo))/(|x_1-x_2|)$. Again, am I wrong?

[gugo82]

"elgiovo":1. (*) defines a bounded Lipschitz continuous function for each $n$.

$f_n(x)=int_(RR)chi_n(x-y)f(y)"d"y=int_(RR)f(x-y)chi_n(y)"d"y=n/2 int_(-1/n)^(1/n)f(x-y)"d"y$

by commutative property of convolution and a direct calculation.
The last quantity is finite $forall x$, because $f in L_("loc")^1(RR)$, so $f_n$ is bounded.

Mmmm this sounds strange... For:

$f(x):=\{(1/x, " if " x!=0),(0, " if " x=0):}$

is finite for each $x\in RR$ but it's not bounded.
Do you mean that the last member has a least upper bound in absolute value?

"elgiovo":For $f_n$ to be Lipschitz, we need

$|f_n(x_1)-f_n(x_2)|<=C|x_1-x_2|$

for some constant $C
$|f_n(x_1)-f_n(x_2)|=|n/2 int_(-1/n)^(1/n)f(x_1-y)"d"y-n/2 int_(-1/n)^(1/n)f(x_2-y)"d"y|=n/2|int_(-1/n)^(1/n)f(x_1-y)-f(x_2-y)"d"y|<=n/2 | ||f||_(oo) int_(-1/n)^(1/n)"d"y|=2 ||f||_(oo)$,

so $f_n$ is locally Lipschitz continuous.

Sorry, I can't understand. Do you mean that a bounded function is always locally Lipschitz?

[elgiovo]

1. (*) defines a bounded Lipschitz continuous function for each $n$.

$f_n(x)=int_(RR)chi_n(x-y)f(y)"d"y=int_(RR)f(x-y)chi_n(y)"d"y=n/2 int_(-1/n)^(1/n)f(x-y)"d"y$

by commutative property of convolution and a direct calculation.
The last quantity is finite $forall x$, because $f in L_("loc")^1(RR)$, so $f_n$ is bounded. For $f_n$ to be Lipschitz, we need

$|f_n(x_1)-f_n(x_2)|<=C|x_1-x_2|$

for some constant $C
$|f_n(x_1)-f_n(x_2)|=|n/2 int_(-1/n)^(1/n)f(x_1-y)"d"y-n/2 int_(-1/n)^(1/n)f(x_2-y)"d"y|=n/2|int_(-1/n)^(1/n)f(x_1-y)-f(x_2-y)"d"y|<=n/2 | "2||f||"_(oo) int_(-1/n)^(1/n)"d"y|=2 ||f||_(oo)$

so $f_n$ is locally Lipschitz continuous.