Mixing
Convolution operator has norm 1 with respect to $L^2$ norm
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Let $C_c(mathbbR)$ be the following:
$$C_c = geq T$$
Let $T_n in L(C_c(mathbbR))$ be a linear operator such that:
$$T_n u = delta_n *u, forall u in C_c(mathbbR),$$
where
$$
delta_n(t)= begincases
n^2(t+1/n) & -1/n leq t leq 0 \
-n^2(t-1/n) & 0 < t leq 1/n \
0 & textelsewhere
endcases
$$
I have to prove that with respect to the 2-norm, the operator $T_n$ has $|T_n| = 1$ and I have succeeded in proving that $|T_nu|_2 leq |u|_2, forall u in C_c(mathbbR).$
Now, I remained to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.
norm convolution
asked Jul 24 at 14:56
giovanni_13
566
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up vote
2
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favorite
Let $C_c(mathbbR)$ be the following:
$$C_c = geq T$$
Let $T_n in L(C_c(mathbbR))$ be a linear operator such that:
$$T_n u = delta_n *u, forall u in C_c(mathbbR),$$
where
$$
delta_n(t)= begincases
n^2(t+1/n) & -1/n leq t leq 0 \
-n^2(t-1/n) & 0 < t leq 1/n \
0 & textelsewhere
endcases
$$
I have to prove that with respect to the 2-norm, the operator $T_n$ has $|T_n| = 1$ and I have succeeded in proving that $|T_nu|_2 leq |u|_2, forall u in C_c(mathbbR).$
Now, I remained to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.
norm convolution
asked Jul 24 at 14:56
giovanni_13
566
1
You don't have to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$. You have to prove that it exists a sequence $(u_n)$ in $C_c(mathbbR)$ with $Vert u_n Vert_2 = 1$ such that $lim Vert T_n u_n Vert_2 = 1$.
– mathcounterexamples.net
Jul 24 at 15:09
@mathcounterexamples.net can you explain me the theory behind it? or give me any reference? And the last limit you wrote, the limit is just for $u_n$ I guess and not for $T_n$. Therefore it's like the $lim_mtoinfty ||T_nu_m||_2$
– giovanni_13
Jul 24 at 15:37
You’re right. The limit is just for $u_m$.
– mathcounterexamples.net
Jul 24 at 16:08
@mathcounterexamples.net any ideas on how this sequence can be?
– giovanni_13
Jul 24 at 16:11
See my hint below.
– mathcounterexamples.net
Jul 24 at 16:33
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $C_c(mathbbR)$ be the following:
$$C_c = geq T$$
Let $T_n in L(C_c(mathbbR))$ be a linear operator such that:
$$T_n u = delta_n *u, forall u in C_c(mathbbR),$$
where
$$
delta_n(t)= begincases
n^2(t+1/n) & -1/n leq t leq 0 \
-n^2(t-1/n) & 0 < t leq 1/n \
0 & textelsewhere
endcases
$$
I have to prove that with respect to the 2-norm, the operator $T_n$ has $|T_n| = 1$ and I have succeeded in proving that $|T_nu|_2 leq |u|_2, forall u in C_c(mathbbR).$
Now, I remained to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.
norm convolution
asked Jul 24 at 14:56
giovanni_13
566
Let $C_c(mathbbR)$ be the following:
$$C_c = geq T$$
Let $T_n in L(C_c(mathbbR))$ be a linear operator such that:
$$T_n u = delta_n *u, forall u in C_c(mathbbR),$$
where
$$
delta_n(t)= begincases
n^2(t+1/n) & -1/n leq t leq 0 \
-n^2(t-1/n) & 0 < t leq 1/n \
0 & textelsewhere
endcases
$$
I have to prove that with respect to the 2-norm, the operator $T_n$ has $|T_n| = 1$ and I have succeeded in proving that $|T_nu|_2 leq |u|_2, forall u in C_c(mathbbR).$
Now, I remained to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.
norm convolution
asked Jul 24 at 14:56
giovanni_13
566
asked Jul 24 at 14:56
giovanni_13
566
asked Jul 24 at 14:56
giovanni_13
566
asked Jul 24 at 14:56
giovanni_13
566
566
1
You don't have to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$. You have to prove that it exists a sequence $(u_n)$ in $C_c(mathbbR)$ with $Vert u_n Vert_2 = 1$ such that $lim Vert T_n u_n Vert_2 = 1$.
– mathcounterexamples.net
Jul 24 at 15:09
@mathcounterexamples.net can you explain me the theory behind it? or give me any reference? And the last limit you wrote, the limit is just for $u_n$ I guess and not for $T_n$. Therefore it's like the $lim_mtoinfty ||T_nu_m||_2$
– giovanni_13
Jul 24 at 15:37
You’re right. The limit is just for $u_m$.
– mathcounterexamples.net
Jul 24 at 16:08
@mathcounterexamples.net any ideas on how this sequence can be?
– giovanni_13
Jul 24 at 16:11
See my hint below.
– mathcounterexamples.net
Jul 24 at 16:33
add a comment |Â
1
You don't have to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$. You have to prove that it exists a sequence $(u_n)$ in $C_c(mathbbR)$ with $Vert u_n Vert_2 = 1$ such that $lim Vert T_n u_n Vert_2 = 1$.
– mathcounterexamples.net
Jul 24 at 15:09
@mathcounterexamples.net can you explain me the theory behind it? or give me any reference? And the last limit you wrote, the limit is just for $u_n$ I guess and not for $T_n$. Therefore it's like the $lim_mtoinfty ||T_nu_m||_2$
– giovanni_13
Jul 24 at 15:37
You’re right. The limit is just for $u_m$.
– mathcounterexamples.net
Jul 24 at 16:08
@mathcounterexamples.net any ideas on how this sequence can be?
– giovanni_13
Jul 24 at 16:11
See my hint below.
– mathcounterexamples.net
Jul 24 at 16:33
1
1
You don't have to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$. You have to prove that it exists a sequence $(u_n)$ in $C_c(mathbbR)$ with $Vert u_n Vert_2 = 1$ such that $lim Vert T_n u_n Vert_2 = 1$.
– mathcounterexamples.net
Jul 24 at 15:09
You don't have to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$. You have to prove that it exists a sequence $(u_n)$ in $C_c(mathbbR)$ with $Vert u_n Vert_2 = 1$ such that $lim Vert T_n u_n Vert_2 = 1$.
– mathcounterexamples.net
Jul 24 at 15:09
@mathcounterexamples.net can you explain me the theory behind it? or give me any reference? And the last limit you wrote, the limit is just for $u_n$ I guess and not for $T_n$. Therefore it's like the $lim_mtoinfty ||T_nu_m||_2$
– giovanni_13
Jul 24 at 15:37
@mathcounterexamples.net can you explain me the theory behind it? or give me any reference? And the last limit you wrote, the limit is just for $u_n$ I guess and not for $T_n$. Therefore it's like the $lim_mtoinfty ||T_nu_m||_2$
– giovanni_13
Jul 24 at 15:37
You’re right. The limit is just for $u_m$.
– mathcounterexamples.net
Jul 24 at 16:08
You’re right. The limit is just for $u_m$.
– mathcounterexamples.net
Jul 24 at 16:08
@mathcounterexamples.net any ideas on how this sequence can be?
– giovanni_13
Jul 24 at 16:11
@mathcounterexamples.net any ideas on how this sequence can be?
– giovanni_13
Jul 24 at 16:11
See my hint below.
– mathcounterexamples.net
Jul 24 at 16:33
See my hint below.
– mathcounterexamples.net
Jul 24 at 16:33
add a comment |Â
2 Answers
2
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up vote
2
down vote
Hint
Consider the function
$$v_m(x)=begincases
0 & x le -frac1m\
1+mx& -frac1m le x le 0\
1 & 0le x le m\
m^2-mx+1& mle x le m+frac1m\
0 & m+frac1m
endcases$$
And the function $u_m =a_m v_m $ with $a_m$ chosen such that $Vert u_m Vert_2=1$. You’ll be able to prove that
$$limlimits_m to infty Vert T_n u_mVert_2 =1.$$
Notice in particular that:
$$T_nv_m(x)=
begincases
0 & x le -frac1n - frac1m\
1 & frac1n le x le m -frac1n\
0 & m+frac1n + frac1m le x
endcases$$ and $0 le T_nv_m(x) le 1$ elsewhere.
For example for $frac1n le x le m -frac1n$:
$$T_nv_m(x)=int_-infty^infty delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t) dt=1$$
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
 |Â
show 5 more comments
up vote
0
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accepted
Probable solution:
What we are interested in is proving that:
$$lim_mtoinfty |T_nu_m|_2 = 1$$
If we write $u_m(t) = a_mv_m(t) $, where $v_m(t)$ defined as the hint by @mathcounterexamples.net, we can substitute in the limit (for simplicity we use $|T_nu_m|_2^2$):
$$lim_mtoinfty |T_nu_m|_2^2 = lim_mtoinfty a_m^2|T_nv_m|_2^2, $$
where $a_m^2 = frac1v_m = frac3m3m^2+2$
Then, according to what @mathcounterexamples.net pointed out, the integral can be split as:
$$|T_nv_m|_2^2 = int_-infty^infty|(delta_n*v_m)(t)|^2dt $$
$$ = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/n(delta_n*v_m)^2(t)dt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt $$
Plugging it back in the limit:
$$lim_mtoinfty a_m^2bigg(int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dtbigg)$$
But in the first and in the third integral:
$$0 leq (delta_n*v_m)(t) leq 1 Rightarrow 0 leq (delta_n*v_m)^2(t) leq 1$$
Therefore, for the first integral:
$$lim_mtoinfty a_m^2int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [-1/n-1/m,1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
And for the third integral as well:
$$lim_mtoinfty a_m^2int_m-1/n^m+1/m+1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [m-1/n,m+1/m+1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
Finally we are left with:
$$lim_mtoinfty a_m^2bigg(int_1/n^m-1/ndtbigg) = lim_mtoinfty a_m^2(m-2/n) = lim_mtoinftyfrac3m3m^2+2(m-2/n) = 1 $$
answered Jul 26 at 13:13
giovanni_13
566
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint
Consider the function
$$v_m(x)=begincases
0 & x le -frac1m\
1+mx& -frac1m le x le 0\
1 & 0le x le m\
m^2-mx+1& mle x le m+frac1m\
0 & m+frac1m
endcases$$
And the function $u_m =a_m v_m $ with $a_m$ chosen such that $Vert u_m Vert_2=1$. You’ll be able to prove that
$$limlimits_m to infty Vert T_n u_mVert_2 =1.$$
Notice in particular that:
$$T_nv_m(x)=
begincases
0 & x le -frac1n - frac1m\
1 & frac1n le x le m -frac1n\
0 & m+frac1n + frac1m le x
endcases$$ and $0 le T_nv_m(x) le 1$ elsewhere.
For example for $frac1n le x le m -frac1n$:
$$T_nv_m(x)=int_-infty^infty delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t) dt=1$$
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
 |Â
show 5 more comments
up vote
2
down vote
Hint
Consider the function
$$v_m(x)=begincases
0 & x le -frac1m\
1+mx& -frac1m le x le 0\
1 & 0le x le m\
m^2-mx+1& mle x le m+frac1m\
0 & m+frac1m
endcases$$
And the function $u_m =a_m v_m $ with $a_m$ chosen such that $Vert u_m Vert_2=1$. You’ll be able to prove that
$$limlimits_m to infty Vert T_n u_mVert_2 =1.$$
Notice in particular that:
$$T_nv_m(x)=
begincases
0 & x le -frac1n - frac1m\
1 & frac1n le x le m -frac1n\
0 & m+frac1n + frac1m le x
endcases$$ and $0 le T_nv_m(x) le 1$ elsewhere.
For example for $frac1n le x le m -frac1n$:
$$T_nv_m(x)=int_-infty^infty delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t) dt=1$$
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
 |Â
show 5 more comments
up vote
2
down vote
up vote
2
down vote
Hint
Consider the function
$$v_m(x)=begincases
0 & x le -frac1m\
1+mx& -frac1m le x le 0\
1 & 0le x le m\
m^2-mx+1& mle x le m+frac1m\
0 & m+frac1m
endcases$$
And the function $u_m =a_m v_m $ with $a_m$ chosen such that $Vert u_m Vert_2=1$. You’ll be able to prove that
$$limlimits_m to infty Vert T_n u_mVert_2 =1.$$
Notice in particular that:
$$T_nv_m(x)=
begincases
0 & x le -frac1n - frac1m\
1 & frac1n le x le m -frac1n\
0 & m+frac1n + frac1m le x
endcases$$ and $0 le T_nv_m(x) le 1$ elsewhere.
For example for $frac1n le x le m -frac1n$:
$$T_nv_m(x)=int_-infty^infty delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t) dt=1$$
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
Hint
Consider the function
$$v_m(x)=begincases
0 & x le -frac1m\
1+mx& -frac1m le x le 0\
1 & 0le x le m\
m^2-mx+1& mle x le m+frac1m\
0 & m+frac1m
endcases$$
And the function $u_m =a_m v_m $ with $a_m$ chosen such that $Vert u_m Vert_2=1$. You’ll be able to prove that
$$limlimits_m to infty Vert T_n u_mVert_2 =1.$$
Notice in particular that:
$$T_nv_m(x)=
begincases
0 & x le -frac1n - frac1m\
1 & frac1n le x le m -frac1n\
0 & m+frac1n + frac1m le x
endcases$$ and $0 le T_nv_m(x) le 1$ elsewhere.
For example for $frac1n le x le m -frac1n$:
$$T_nv_m(x)=int_-infty^infty delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t)v_m(t) dt = int_x-frac1n^x+frac1n delta_n(x-t) dt=1$$
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
edited Jul 25 at 16:42
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
answered Jul 24 at 16:32
mathcounterexamples.net
23.9k21653
23.9k21653
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
 |Â
show 5 more comments
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
But is this together with the first part where I proved that $||T_nu||_2 leq ||u||_2$ or this alone is enough?
– giovanni_13
Jul 25 at 7:43
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
This is together with what your prove.
– mathcounterexamples.net
Jul 25 at 7:47
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I tried to compute the Fourier transform first in order to check if, after the right coefficient $a_m$, what's the form of $lim_mtoinfty ||T_nu_m||_2 = lim_mtoinfty ||mathfrakF(T_nu_m)||_2 = lim_mtoinfty ||mathfrakF(T_n)mathfrakF(u_m)||_2$, but I got that $mathfrakF(v_m)(iw) = (m/w^2)(e^-iwm - e^-iw/m - e^-iw(m+1/m))$. Is that correct?
– giovanni_13
Jul 25 at 11:54
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
I don't think that you need to go through Fourier transform. I updated my answer to provide you with more details.
– mathcounterexamples.net
Jul 25 at 13:55
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
Actually, I think that you are not completely right. In fact, what you wrote is true for $n=1$, i.e., $T_1v_m$. Whereas, when $n neq 1$, then $T_nv_m = (2n-1)/n^2$, for $1/n leq x leq m-1/n$
– giovanni_13
Jul 25 at 16:11
 |Â
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Probable solution:
What we are interested in is proving that:
$$lim_mtoinfty |T_nu_m|_2 = 1$$
If we write $u_m(t) = a_mv_m(t) $, where $v_m(t)$ defined as the hint by @mathcounterexamples.net, we can substitute in the limit (for simplicity we use $|T_nu_m|_2^2$):
$$lim_mtoinfty |T_nu_m|_2^2 = lim_mtoinfty a_m^2|T_nv_m|_2^2, $$
where $a_m^2 = frac1v_m = frac3m3m^2+2$
Then, according to what @mathcounterexamples.net pointed out, the integral can be split as:
$$|T_nv_m|_2^2 = int_-infty^infty|(delta_n*v_m)(t)|^2dt $$
$$ = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/n(delta_n*v_m)^2(t)dt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt $$
Plugging it back in the limit:
$$lim_mtoinfty a_m^2bigg(int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dtbigg)$$
But in the first and in the third integral:
$$0 leq (delta_n*v_m)(t) leq 1 Rightarrow 0 leq (delta_n*v_m)^2(t) leq 1$$
Therefore, for the first integral:
$$lim_mtoinfty a_m^2int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [-1/n-1/m,1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
And for the third integral as well:
$$lim_mtoinfty a_m^2int_m-1/n^m+1/m+1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [m-1/n,m+1/m+1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
Finally we are left with:
$$lim_mtoinfty a_m^2bigg(int_1/n^m-1/ndtbigg) = lim_mtoinfty a_m^2(m-2/n) = lim_mtoinftyfrac3m3m^2+2(m-2/n) = 1 $$
answered Jul 26 at 13:13
giovanni_13
566
add a comment |Â
up vote
0
down vote
accepted
Probable solution:
What we are interested in is proving that:
$$lim_mtoinfty |T_nu_m|_2 = 1$$
If we write $u_m(t) = a_mv_m(t) $, where $v_m(t)$ defined as the hint by @mathcounterexamples.net, we can substitute in the limit (for simplicity we use $|T_nu_m|_2^2$):
$$lim_mtoinfty |T_nu_m|_2^2 = lim_mtoinfty a_m^2|T_nv_m|_2^2, $$
where $a_m^2 = frac1v_m = frac3m3m^2+2$
Then, according to what @mathcounterexamples.net pointed out, the integral can be split as:
$$|T_nv_m|_2^2 = int_-infty^infty|(delta_n*v_m)(t)|^2dt $$
$$ = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/n(delta_n*v_m)^2(t)dt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt $$
Plugging it back in the limit:
$$lim_mtoinfty a_m^2bigg(int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dtbigg)$$
But in the first and in the third integral:
$$0 leq (delta_n*v_m)(t) leq 1 Rightarrow 0 leq (delta_n*v_m)^2(t) leq 1$$
Therefore, for the first integral:
$$lim_mtoinfty a_m^2int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [-1/n-1/m,1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
And for the third integral as well:
$$lim_mtoinfty a_m^2int_m-1/n^m+1/m+1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [m-1/n,m+1/m+1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
Finally we are left with:
$$lim_mtoinfty a_m^2bigg(int_1/n^m-1/ndtbigg) = lim_mtoinfty a_m^2(m-2/n) = lim_mtoinftyfrac3m3m^2+2(m-2/n) = 1 $$
answered Jul 26 at 13:13
giovanni_13
566
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Probable solution:
What we are interested in is proving that:
$$lim_mtoinfty |T_nu_m|_2 = 1$$
If we write $u_m(t) = a_mv_m(t) $, where $v_m(t)$ defined as the hint by @mathcounterexamples.net, we can substitute in the limit (for simplicity we use $|T_nu_m|_2^2$):
$$lim_mtoinfty |T_nu_m|_2^2 = lim_mtoinfty a_m^2|T_nv_m|_2^2, $$
where $a_m^2 = frac1v_m = frac3m3m^2+2$
Then, according to what @mathcounterexamples.net pointed out, the integral can be split as:
$$|T_nv_m|_2^2 = int_-infty^infty|(delta_n*v_m)(t)|^2dt $$
$$ = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/n(delta_n*v_m)^2(t)dt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt $$
Plugging it back in the limit:
$$lim_mtoinfty a_m^2bigg(int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dtbigg)$$
But in the first and in the third integral:
$$0 leq (delta_n*v_m)(t) leq 1 Rightarrow 0 leq (delta_n*v_m)^2(t) leq 1$$
Therefore, for the first integral:
$$lim_mtoinfty a_m^2int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [-1/n-1/m,1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
And for the third integral as well:
$$lim_mtoinfty a_m^2int_m-1/n^m+1/m+1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [m-1/n,m+1/m+1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
Finally we are left with:
$$lim_mtoinfty a_m^2bigg(int_1/n^m-1/ndtbigg) = lim_mtoinfty a_m^2(m-2/n) = lim_mtoinftyfrac3m3m^2+2(m-2/n) = 1 $$
answered Jul 26 at 13:13
giovanni_13
566
Probable solution:
What we are interested in is proving that:
$$lim_mtoinfty |T_nu_m|_2 = 1$$
If we write $u_m(t) = a_mv_m(t) $, where $v_m(t)$ defined as the hint by @mathcounterexamples.net, we can substitute in the limit (for simplicity we use $|T_nu_m|_2^2$):
$$lim_mtoinfty |T_nu_m|_2^2 = lim_mtoinfty a_m^2|T_nv_m|_2^2, $$
where $a_m^2 = frac1v_m = frac3m3m^2+2$
Then, according to what @mathcounterexamples.net pointed out, the integral can be split as:
$$|T_nv_m|_2^2 = int_-infty^infty|(delta_n*v_m)(t)|^2dt $$
$$ = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/n(delta_n*v_m)^2(t)dt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt = int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dt $$
Plugging it back in the limit:
$$lim_mtoinfty a_m^2bigg(int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt + int_1/n^m-1/ndt + int_m-1/n^1/n+1/m+m(delta_n*v_m)^2(t)dtbigg)$$
But in the first and in the third integral:
$$0 leq (delta_n*v_m)(t) leq 1 Rightarrow 0 leq (delta_n*v_m)^2(t) leq 1$$
Therefore, for the first integral:
$$lim_mtoinfty a_m^2int_-1/n-1/m^1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [-1/n-1/m,1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
And for the third integral as well:
$$lim_mtoinfty a_m^2int_m-1/n^m+1/m+1/n(delta_n*v_m)^2(t)dt leq lim_mtoinfty a_m^2bigg(sup_t in [m-1/n,m+1/m+1/n](delta_n*v_m)^2(t)bigg)^2(2/n+1/m) = lim_mtoinfty a_m^2(2/n+1/m) = 0$$
Finally we are left with:
$$lim_mtoinfty a_m^2bigg(int_1/n^m-1/ndtbigg) = lim_mtoinfty a_m^2(m-2/n) = lim_mtoinftyfrac3m3m^2+2(m-2/n) = 1 $$
answered Jul 26 at 13:13
giovanni_13
566
answered Jul 26 at 13:13
giovanni_13
566
answered Jul 26 at 13:13
giovanni_13
566
answered Jul 26 at 13:13
giovanni_13
566
566
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1
You don't have to prove that $exists text u in C_c(mathbbR) text s.t. |T_nu|_2 = |u|_2$. You have to prove that it exists a sequence $(u_n)$ in $C_c(mathbbR)$ with $Vert u_n Vert_2 = 1$ such that $lim Vert T_n u_n Vert_2 = 1$.
– mathcounterexamples.net
Jul 24 at 15:09
@mathcounterexamples.net can you explain me the theory behind it? or give me any reference? And the last limit you wrote, the limit is just for $u_n$ I guess and not for $T_n$. Therefore it's like the $lim_mtoinfty ||T_nu_m||_2$
– giovanni_13
Jul 24 at 15:37
You’re right. The limit is just for $u_m$.
– mathcounterexamples.net
Jul 24 at 16:08
@mathcounterexamples.net any ideas on how this sequence can be?
– giovanni_13
Jul 24 at 16:11
See my hint below.
– mathcounterexamples.net
Jul 24 at 16:33