Integral Equation & Boundary Values Problem
Let $a
It's known that each $K in C([a,b]^2)$ can be used to define a compact linear operator $T$ of $L^p([a,b])$ into $L^q([a,b])$ (with $p,q in ]1,+oo[$ s.t. $1/p+1/q=1$) by putting:
(*) $quad AAu in L^p([a,b]), quad Tu(x)=\int_a^bK(x,y)*u(y)" d"y quad$,
where $"d"y$ stands for the standard Lebesgue measure on $[a,b]$. (No need to proof; it's just a prerequisite.)
On the other hand, once assigned a linear differential operator of the second order $L:C^2([a,b]) to C([a,b])$, a function $phi in C([a,b])$ and a couple of real numbers $u_a,u_b$, a problem of the following type:
(**) $quad \{(Lu(x)=phi(x)),(u(a)=u_a),(u(b)=u_b):}$
is called boundary values problem (or b.v.p. for short) for the operator $L$; if $u_a=0=u_b$ we say that (**) is a b.v.p. with homogeneous boundary conditions (or a b.v.p.h.).
***
Put $AA t in RR, (t)^+=max{0,t}$, choose $K(x,y)=x/(2pi)(2pi-y)-(x-y)^+$ so that $K in C([0,2pi]^2)$, define the compact operator $T$ in $L^2$ into itself as in (*) and let $mu$ be any real number.
Prove that:
1) each solution of the homogeneous equation $u-muTu=omega$ ($omega$ being the function equal to zero a.e. in $[0,2pi]$) is of class $C^2$ (hint: first prove $u in C$; at the beginning you could find useful Fubini's Theorem to rearrange two double integrals...);
2) each solution of the above-mentioned equation also solves a b.v.p.h. in $[0,2pi]$ for a linear ODE of the second order with constant coefficients;
3) each solution of the b.v.p.h. in 2) solves the homogeneous equation in 1) (therefore the integral equation and the b.v.p.h. are equivalent);
4) find regular and singular values for the complete equation $u=f+mu*Tu$ (hint: remember the Fredholm Alternative; use the representation formula for the solutions of the b.v.p.h. equivalent to our homogeneous equation);
5) find the eigenvalues of the compact operator $T$.
(*) $quad AAu in L^p([a,b]), quad Tu(x)=\int_a^bK(x,y)*u(y)" d"y quad$,
where $"d"y$ stands for the standard Lebesgue measure on $[a,b]$. (No need to proof; it's just a prerequisite.)
On the other hand, once assigned a linear differential operator of the second order $L:C^2([a,b]) to C([a,b])$, a function $phi in C([a,b])$ and a couple of real numbers $u_a,u_b$, a problem of the following type:
(**) $quad \{(Lu(x)=phi(x)),(u(a)=u_a),(u(b)=u_b):}$
is called boundary values problem (or b.v.p. for short) for the operator $L$; if $u_a=0=u_b$ we say that (**) is a b.v.p. with homogeneous boundary conditions (or a b.v.p.h.).
***
Put $AA t in RR, (t)^+=max{0,t}$, choose $K(x,y)=x/(2pi)(2pi-y)-(x-y)^+$ so that $K in C([0,2pi]^2)$, define the compact operator $T$ in $L^2$ into itself as in (*) and let $mu$ be any real number.
Prove that:
1) each solution of the homogeneous equation $u-muTu=omega$ ($omega$ being the function equal to zero a.e. in $[0,2pi]$) is of class $C^2$ (hint: first prove $u in C$; at the beginning you could find useful Fubini's Theorem to rearrange two double integrals...);
2) each solution of the above-mentioned equation also solves a b.v.p.h. in $[0,2pi]$ for a linear ODE of the second order with constant coefficients;
3) each solution of the b.v.p.h. in 2) solves the homogeneous equation in 1) (therefore the integral equation and the b.v.p.h. are equivalent);
4) find regular and singular values for the complete equation $u=f+mu*Tu$ (hint: remember the Fredholm Alternative; use the representation formula for the solutions of the b.v.p.h. equivalent to our homogeneous equation);
5) find the eigenvalues of the compact operator $T$.
Risposte
"Gugo82":
4) find regular and singular values for the complete equation $u=f+mu*Tu$ (hint: remember the Fredholm Alternative; use the representation formula for the solutions of the h.b.v.p. equivalent to our homogeneous integral equation);
Because of the Fredholm Alternative, in order to find singular values for the equation $u=f+mu*Tu$ we have to determine the set of the scalars $mu$ s.t. the homogeneous equation $u-mu*Tu=omega$ has non-trivial solutions.
From what we proved above it follows that $u-mu*Tu=omega$ has non-trivial solutions if and only if $\{(u''+mu*u=0),(u(0)=0=u(2pi)):}$ does.
The latter h.b.v.p. is easy to solve: infact we have to consider only the following three eventualities:
a) $mu<0$: for these values of $mu$, the ODE $u''+mu*u=0$ has general integral equal to $u(x)=c_1*e^(-sqrt|mu| x)+c_2*e^(sqrt|mu| x)$ hence the h.b.v.p. possesses only the trivial solution $u=0$ in $[0,2pi]$;
b) $mu=0$: the ODE $u''=0$ has general integral equal to $u(x)=c_1+c_2*x$ hence our h.b.v.p. has again only the trivial solution in $[0,2pi]$;
c) $mu>0$: the ODE $u''+mu*u=0$ has general integral equal to $u(x)=c_1*cos(sqrtmu x)+c_2*sin(sqrtmu x)$ therefore the solution of our h.b.v.p. has to satisfy the conditions:
$\{(c_1=0),(c_1*cos(sqrtmu 2pi)+c_2*sin(sqrtmu 2pi)=0):} quad hArr quad \{(c_1=0),(c_2*sin(sqrtmu 2pi)=0):} quad$;
it's easily seen that the latter linear sistem of equations has non-trivial solutions iff $sin(sqrtmu 2pi)=0$, i.e. if and only $mu$ satisfies the condition $mu=n^2/4 quad$ for some $n in NN$.
The statements in a-b-c) imply that the singular values of our integral equation are the elements of the sequence $S={n^2/4}_(n in NN)$, so that the set $RR-S$ contains the whole regular values of the equation $u=f+mu*Tu$.
"Gugo82":
5) find the eigenvalues of the compact operator $T$.
As long as the non-zero eigenvalues of $T$ are the inverse of our integral equation's singular values, the $n$-th eigenvalue of $T$ is:
$lambda_n=4/n^2 quad$ ($n in NN$).
Therefore ${4/n^2}_(n in NN) =sigma_P(T)\setminus \{0\}$; actually I don't know if $0 in sigma_P(T)$ or $0 in sigma(T)\setminus sigma_P(T)$.
"Gugo82":
2) each solution of the above-mentioned equation also solves a b.v.p.h. in $[0,2pi]$ for a linear ODE of the second order with constant coefficients;
We can differentiate both r.h. and l.h. sides of (***) and have:
$u'(x)=mu/pi*\int_0^(2pi) \int_0^t u(y) " d"y" d"t-mu*\int_0^x u(y)" d"y quad =>$
$quad => quad u''(x)=-mu*u(x) quad hArr$
$quad hArr quad u''(x)+mu*u(x)=0$
so that $u$ solves an homogeneous linear ODE with constant coefficients; moreover we find:
$u(0)=0-mu*\int_0^0 \int_0^t u(y)" d"y" d"t=0 quad$ and $quad u(2pi)=mu*\int_0^(2pi) \int_0^(2pi) u(y)" d"y" d"t-mu*\int_0^(2pi) \int_0^(2pi) u(y)" d"y" d"t=0$.
Therefore a solution $u$ of the integral equation $u-mu*Tu=omega$ also solves the h.b.v.p.:
$\{(u''+mu*u=0),(u(0)=0=u(2pi)) :}$
in $[0,2pi]$.
"Gugo82":
3) each solution of the b.v.p.h. in 2) solves the homogeneous equation in 1) (therefore the integral equation and the b.v.p.h. are equivalent);
Let $u in C^2([0,2pi])$ solve:
$\{(u''+mu*u=0),(u(0)=0=u(2pi)) :} quad$;
integrating two times both sides of the previous ODE with start-point $0$ and end-point $x in [0,2pi]$ we find:
$u'(x)=u'(0)-mu*\int_0^x u(y)" d"y quad =>$
$quad => quad u(x)=u'(0)*x-mu \int_0^x \int_0^t u(y)" d"y" d"t$.
The condition $u(2pi)=0$ implies:
$u'(0)=mu/(2pi)*\int_0^(2pi) \int_0^t u(y)" d"y" d"t$
so that $u$ solves the integral equation:
$u(x)=mu/(2pi)*x*\int_0^(2pi) \int_0^t u(y)" d"y" d"t-mu*\int_0^x \int_0^t u(y)" d"y" d"t$
which is our (*) up to the modifications due to an application of Fubini's theorem.
It's been a month since I wrote this thread's first post: even if Thomas tried to solve the first part of the problem, no one has given a complete answer.
So I give you my solution: I hope someone like it.
Remembering (*) and the very definition of $K$, we have:
(**) $quad u(x)=mu/(2pi)*x*\int_0^(2pi) (2pi-y)*u(y)" d"y-mu*\int_0^x(x-y)*u(y)" d"y quad$ in the sense of $L^2$ (i.e. almost everywhere in $[a,b]$);
since $2pi-y=\int_y^(2pi)" d"t$ and $x-y=\int_y^x" d"t$, the integrals in the previous formula's second member can be written as:
1.1) $quad \int_0^(2pi) (2pi-y)*u(y)" d"y=\int_0^(2pi) (\int_y^(2pi)" d"t)*u(y)" d"y$
1.2) $quad \int_0^x(x-y)*u(y)" d"y=\int_0^x(\int_y^x" d"t)*u(y)" d"y$.
The latter two integrals are extended to the two triangles $T={(t,y) in RR^2:quad 0le tle yle 2pi}, T_x={(t,y) in RR^2:quad 0le tle yle x} subset [0,2pi]^2$ and they're absolutely convergent because of the l.h.sides of 1.1-2): therefore we can apply Fubini's theorem and write:
$\int_0^(2pi) (2pi-y)*u(y)" d"y=\int \int_T u(y)" d"t" d"y quad$ and $quad \int_0^x(x-y)*u(y)" d"y=\int \int_(T_x) u(y)" d"t" d"y$
hence, integrating by orizontals, we have:
1.3) $quad \int_0^(2pi) (2pi-y)*u(y)" d"y=\int_0^(2pi) (\int_0^t u(y)" d"y)" d"t$;
1.4) $quad \int_0^x(x-y)*u(y)" d"y=\int_0^x(\int_0^t u(y)" d"y)" d"t$.
A substitution of 1.3-4) in (**) leads to:
(***) $quad u(x)=mu/(2pi)*x*\int_0^(2pi) (\int_0^t u(y)" d"y)" d"t-mu*\int_0^x(\int_0^t u(y)" d"y)" d"t$.
The first summand in the r.h.side is a linear function of $x$, hence it's of class $C^oo$ in $[0,2pi]$; the second summand is the integral function with start-point $0$ of $eta(x)=\int_0^x u(y)" d"y$: since $u$ is Lebesgue measurable, $eta$ is absolutely continuous and a fortiori continuous in $[0,2pi]$, therefore $\int_0^x(\int_0^t u(y)" d"y)" d"t=\int_0^x eta(t)" d"t$ is of class $C^1$ in $[0,2pi]$.
It follows that $u in C^1([0,2pi])$; but this fact implies that $eta$ is of class $C^1$ (because it's the integral function of $u$, which is continuous), therefore $u$ is the sum of a $C^oo$-function and a $C^2$-function and finally $u in C^2([0,2pi])$.
So I give you my solution: I hope someone like it.
"Gugo82":
Let $a It's known that each $K in C([a,b]^2)$ can be used to define a compact linear operator $T$ of $L^p([a,b])$ into $L^q([a,b])$ (with $p,q in ]1,+oo[$ s.t. $1/p+1/q=1$) by putting:
(*) $quad AAu in L^p([a,b]), quad Tu(x)=\int_a^bK(x,y)*u(y)" d"y quad$,
where $"d"y$ stands for the standard Lebesgue measure on $[a,b]$. [...]
Put $AA t in RR, (t)^+=max{0,t}$, choose $K(x,y)=x/(2pi)(2pi-y)-(x-y)^+$ so that $K in C([0,2pi]^2)$, define the compact operator $T$ in $L^2$ into itself as in (*) and let $mu$ be any real number.
Prove that:
1) each solution of the homogeneous equation $u-muTu=omega$ ($omega$ being the function equal to zero a.e. in $[0,2pi]$) is of class $C^2$ (hint: first prove $u in C$; at the beginning you could find useful Fubini's Theorem to rearrange two double integrals...);
Remembering (*) and the very definition of $K$, we have:
(**) $quad u(x)=mu/(2pi)*x*\int_0^(2pi) (2pi-y)*u(y)" d"y-mu*\int_0^x(x-y)*u(y)" d"y quad$ in the sense of $L^2$ (i.e. almost everywhere in $[a,b]$);
since $2pi-y=\int_y^(2pi)" d"t$ and $x-y=\int_y^x" d"t$, the integrals in the previous formula's second member can be written as:
1.1) $quad \int_0^(2pi) (2pi-y)*u(y)" d"y=\int_0^(2pi) (\int_y^(2pi)" d"t)*u(y)" d"y$
1.2) $quad \int_0^x(x-y)*u(y)" d"y=\int_0^x(\int_y^x" d"t)*u(y)" d"y$.
The latter two integrals are extended to the two triangles $T={(t,y) in RR^2:quad 0le tle yle 2pi}, T_x={(t,y) in RR^2:quad 0le tle yle x} subset [0,2pi]^2$ and they're absolutely convergent because of the l.h.sides of 1.1-2): therefore we can apply Fubini's theorem and write:
$\int_0^(2pi) (2pi-y)*u(y)" d"y=\int \int_T u(y)" d"t" d"y quad$ and $quad \int_0^x(x-y)*u(y)" d"y=\int \int_(T_x) u(y)" d"t" d"y$
hence, integrating by orizontals, we have:
1.3) $quad \int_0^(2pi) (2pi-y)*u(y)" d"y=\int_0^(2pi) (\int_0^t u(y)" d"y)" d"t$;
1.4) $quad \int_0^x(x-y)*u(y)" d"y=\int_0^x(\int_0^t u(y)" d"y)" d"t$.
A substitution of 1.3-4) in (**) leads to:
(***) $quad u(x)=mu/(2pi)*x*\int_0^(2pi) (\int_0^t u(y)" d"y)" d"t-mu*\int_0^x(\int_0^t u(y)" d"y)" d"t$.
The first summand in the r.h.side is a linear function of $x$, hence it's of class $C^oo$ in $[0,2pi]$; the second summand is the integral function with start-point $0$ of $eta(x)=\int_0^x u(y)" d"y$: since $u$ is Lebesgue measurable, $eta$ is absolutely continuous and a fortiori continuous in $[0,2pi]$, therefore $\int_0^x(\int_0^t u(y)" d"y)" d"t=\int_0^x eta(t)" d"t$ is of class $C^1$ in $[0,2pi]$.
It follows that $u in C^1([0,2pi])$; but this fact implies that $eta$ is of class $C^1$ (because it's the integral function of $u$, which is continuous), therefore $u$ is the sum of a $C^oo$-function and a $C^2$-function and finally $u in C^2([0,2pi])$.
I continue a bit your calculation:
$K(x,y)=(x/(2pi)(2\pi-y)-(x-y)=y(1-x/(2\pi)).................... 0le yle x le 2pi$
$K(x,y)=(x/(2pi)(2\pi-y)=x(1-y/(2\pi))............... 0le xle y le2pi)$
$K(y,x)=(y/(2pi)(2\pi-x)-(y-x)=x(1-y/(2\pi))........... 0le xle y le 2pi)$
$K(y,x)=(y/(2pi)(2\pi-x)=y(1-x/(2\pi)).......... 0le yle x le2pi)$
now it is evident that it is simmetric, even if this point is of no importance at the moment...or not?
$K(x,y)=(x/(2pi)(2\pi-y)-(x-y)=y(1-x/(2\pi)).................... 0le yle x le 2pi$
$K(x,y)=(x/(2pi)(2\pi-y)=x(1-y/(2\pi))............... 0le xle y le2pi)$
$K(y,x)=(y/(2pi)(2\pi-x)-(y-x)=x(1-y/(2\pi))........... 0le xle y le 2pi)$
$K(y,x)=(y/(2pi)(2\pi-x)=y(1-x/(2\pi)).......... 0le yle x le2pi)$
now it is evident that it is simmetric, even if this point is of no importance at the moment...or not?
"Thomas":
suppose $x>y$, then
$K(x,y)=x(2\pi-y)/(2\pi)-(x-y)=y(1-x/(2\pi))$
$K(y,x)=y(1-x/(2\pi))$
why do u say it isn't simmetric?
The graph of $K$ is not symmetric with respect to the plane $x-y=0$ in $RR^3$; infact:
$K(x,y)=\{(x/(2pi)(2\pi-y)-(x-y), " if "0le yle x le 2pi), (x/(2pi)(2\pi-y), " if " 0le xle y le2pi):}$
$K(y,x)=\{(y/(2pi)(2\pi-x)-(y-x), " if "0le xle y le 2pi), (y/(2pi)(2\pi-x), " if " 0le yle x le2pi):}$
hence $K(x,y)!=K(y,x)$... but maybe I'm wrong. Anyone else can check?
ok ok.... but general results are more interesting (and nicer, but maybe this is a matter of taste!) 
....
furthermore if it's correct the proof of the continuity in the general case is not so complicated.... actually I'd prefer to lose in generality only when it's necessary...
anyway I'll try in the future to end the calculation and see if my way works.....
otherwise I'll try to follow the way u suggest...
bye for now........
ps: it isn't important, but:
suppose $x>y$, then
$K(x,y)=x(2\pi-y)/(2\pi)-(x-y)=y(1-x/(2\pi))$
$K(y,x)=y(1-x/(2\pi))$
why do u say it isn't simmetric?
.... furthermore if it's correct the proof of the continuity in the general case is not so complicated.... actually I'd prefer to lose in generality only when it's necessary...
anyway I'll try in the future to end the calculation and see if my way works.....
otherwise I'll try to follow the way u suggest...
bye for now........
ps: it isn't important, but:
suppose $x>y$, then
$K(x,y)=x(2\pi-y)/(2\pi)-(x-y)=y(1-x/(2\pi))$
$K(y,x)=y(1-x/(2\pi))$
why do u say it isn't simmetric?
Don't know if iterated kernels will work... what I know for sure is that $K$ isn't symmetric (i.e. $K(y,x)!=K(x,y)$).
The proof of the continuity seems good; but I'd like to notice that this proof may become easier if you have a suitable integral rapresentation of $u$ (which can be obtained through Fubini's Theorem, as I said before).
However, mine wasn't a question of "general type"; I was asking about regularity of the solutions of $u(x)=mu\int_a^bK(x,y)u(y)" d"y$ with respect to the above-said kernel and not with respect to any kernel $K in C$.
The proof of the continuity seems good; but I'd like to notice that this proof may become easier if you have a suitable integral rapresentation of $u$ (which can be obtained through Fubini's Theorem, as I said before).
However, mine wasn't a question of "general type"; I was asking about regularity of the solutions of $u(x)=mu\int_a^bK(x,y)u(y)" d"y$ with respect to the above-said kernel and not with respect to any kernel $K in C$.
ok well.... but I wanted to ask if what I wrote is correct.... (at least for the continuity)...
apart from other hints, I proposed an other way to increase the regularity... u think it won't work (actually I've got to compute F(x,z) in order to say that: nevertheless I'm confident ti will end up with the necessary regularity! ) or it is even wrong?
Pheraps u haven't got the same faith, considering u are trying to change my plans!
.... do u see something wrong that i don't see?
apart from other hints, I proposed an other way to increase the regularity... u think it won't work (actually I've got to compute F(x,z) in order to say that: nevertheless I'm confident ti will end up with the necessary regularity! ) or it is even wrong?
Pheraps u haven't got the same faith, considering u are trying to change my plans!
.... do u see something wrong that i don't see?
You actually have to use $K(x,y)=x/(2pi)(2pi-y)-(x-y)^+$ to prove $u in C^2$; in general, it's impossible to prove that $u$ is more regular than continuous for kernels $K$ of general type (even if $K in C^oo$).
Another hint: Keep in mind that:
$u-muTu=omega quad hArr quad u(x)=mux/(2pi)\int_0^(2pi)(2pi-y)u(y)" d"y-mu\int_0^(2pi)(x-y)^+u(y)" d"y=mux/(2pi)\int_0^(2pi)(2pi-y)u(y)" d"y-mu\int_0^x(x-y)u(y)" d"y$;
remember that $2pi-y=\int_y^(2pi)" d"t$ and $x-y=\int_y^x" d"t$; finally use Fubini's Theorem to change the integration order in two double integrals.
Another hint: Keep in mind that:
$u-muTu=omega quad hArr quad u(x)=mux/(2pi)\int_0^(2pi)(2pi-y)u(y)" d"y-mu\int_0^(2pi)(x-y)^+u(y)" d"y=mux/(2pi)\int_0^(2pi)(2pi-y)u(y)" d"y-mu\int_0^x(x-y)u(y)" d"y$;
remember that $2pi-y=\int_y^(2pi)" d"t$ and $x-y=\int_y^x" d"t$; finally use Fubini's Theorem to change the integration order in two double integrals.
I don't know: I've got the impression of "arrampicarmi sugli specchi", saying that with an italian motto...
maybe $F(x,z)$ is not as regular as I think.... Actually I computed this only for $x>z$ at the moment (I realise it only now)... if I had to bet, I would say it is regular enough since the K function is continuous (even if not C^1)...
Now I wait anyway....
maybe $F(x,z)$ is not as regular as I think.... Actually I computed this only for $x>z$ at the moment (I realise it only now)... if I had to bet, I would say it is regular enough since the K function is continuous (even if not C^1)...
Now I wait anyway....
I tried a bit point one...
1) -
---------
Claim: to say that the function is continuous (precisely, it has a continuos representant), we have to fix an arbitrary $x$ and try to say for any $\epsilon$ there exists a $\delta(x)$ s.t.
$|u(y)-u(x)|<\epsilon$
for $|y-x|<\delta(x)$ [1] a.e.
NB: this is not trivial becuase [1] holds only almost everywhere
Trace of a dim (I didn't check that it works, but I hope):
- we can observe that due to the compactness $\delta(x)$ can be chosen such that it is a $\delta$ independent of x.... (recalling the classical proof of uniform continuity).
- then we can modify the function $u$ such that [1] holds everywhere for a dense numerable set $x_n$;
- finally we check that the function so chosen is continuos for every $x$.
------------
so, we can consider
$|u(x+h)-u(x)|<=\mu\int_a^b|K(x+h,y)-K(x,y)||u(y)|dy<=\mu\int_a^b\epsilon|u(y)|dy=\mu\epsilon||u||_1$ [2]
where I've used the uniform continuity of $K(x,y)$ as a funtion in $[a,b]^2$. For any $\epsilon$ I can find a set of values of h close zero in which [2] holds. Furthermore, observe that $u$ is in L^1 thanks to the inclusion between L^p spaces,
Using the lemma, continuity is obtained...
This result is independent of the particular form of $K$.
Actually, I've got some problems at the moment in increasing the regularity of the solution. The way I'd like to follow is this:
$u(x)=\mu\int_a^b \int_a^bK(x,y)K(y,z)u(z)dzdy=\mu\int_a^bdzu(z)\int_a^bK(x,y)K(y,z)dy=\mu\int_a^bdzu(z)F(x,z)$
(where it's been used that $K(x,y)=K(y,x)$ but I don't think this is relevant at the moment)
if by direct computation I show that $F(x,z)$ is differentiable (I think I've computed that!) I could try to apply dominated convergence to conclude that the function is C^1.
Actually, this way I think also that it's been proven that $u$ is $C^\infty$, since the function $F(x,z)$ eventually will be polinomial, according to my calculation...
I'll try to rewrite this in a decent way, once gugo82 will have expressed his opinion on what I've said...
1) -
---------
Claim: to say that the function is continuous (precisely, it has a continuos representant), we have to fix an arbitrary $x$ and try to say for any $\epsilon$ there exists a $\delta(x)$ s.t.
$|u(y)-u(x)|<\epsilon$
for $|y-x|<\delta(x)$ [1] a.e.
NB: this is not trivial becuase [1] holds only almost everywhere
Trace of a dim (I didn't check that it works, but I hope):
- we can observe that due to the compactness $\delta(x)$ can be chosen such that it is a $\delta$ independent of x.... (recalling the classical proof of uniform continuity).
- then we can modify the function $u$ such that [1] holds everywhere for a dense numerable set $x_n$;
- finally we check that the function so chosen is continuos for every $x$.
------------
so, we can consider
$|u(x+h)-u(x)|<=\mu\int_a^b|K(x+h,y)-K(x,y)||u(y)|dy<=\mu\int_a^b\epsilon|u(y)|dy=\mu\epsilon||u||_1$ [2]
where I've used the uniform continuity of $K(x,y)$ as a funtion in $[a,b]^2$. For any $\epsilon$ I can find a set of values of h close zero in which [2] holds. Furthermore, observe that $u$ is in L^1 thanks to the inclusion between L^p spaces,
Using the lemma, continuity is obtained...
This result is independent of the particular form of $K$.
Actually, I've got some problems at the moment in increasing the regularity of the solution. The way I'd like to follow is this:
$u(x)=\mu\int_a^b \int_a^bK(x,y)K(y,z)u(z)dzdy=\mu\int_a^bdzu(z)\int_a^bK(x,y)K(y,z)dy=\mu\int_a^bdzu(z)F(x,z)$
(where it's been used that $K(x,y)=K(y,x)$ but I don't think this is relevant at the moment)
if by direct computation I show that $F(x,z)$ is differentiable (I think I've computed that!) I could try to apply dominated convergence to conclude that the function is C^1.
Actually, this way I think also that it's been proven that $u$ is $C^\infty$, since the function $F(x,z)$ eventually will be polinomial, according to my calculation...
I'll try to rewrite this in a decent way, once gugo82 will have expressed his opinion on what I've said...
hey.... I wanted to say u that I'm very happy to see this exercise!!! actually I miss pure mathematics 
Unluckily in this period I've got exams and I haven't got time...
(why do u see I'm spending time on the forum in the last days? simply because I have to study... it's typical of me: when I've got to do something, I do something else... ... but now I'm terribly running out of time! 
)...
but I promise I'll try to post a clear solution in the future...
.... and of course, if somebody will have anticipated me, I'll try to make it on my own...
bye
Unluckily in this period I've got exams and I haven't got time...
(why do u see I'm spending time on the forum in the last days? simply because I have to study... it's typical of me: when I've got to do something, I do something else...
... but now I'm terribly running out of time! )...but I promise I'll try to post a clear solution in the future...
.... and of course, if somebody will have anticipated me, I'll try to make it on my own...bye
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