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No identity for Convolution of Lebesgue Integrable Functions

0

$begingroup$

I want to show there's no identity for the convolution of two functions. I've found posts that answer this for Riemann integrable functions, but I'm working with the convolution given by $fast g = int_{mathbb{R}^d} f(x-y)g(y) dy$, where $f, g$ are Lebesgue integrable. So, I want to show there does not exist a $deltain{L^1(mathbb{R}^d)}$ such that $fastdelta = f$.

My attempt:

Assume towards a contradiction that there is such a function $delta$. Then we have $fastdelta = f$
and hence,
$$ | fastdelta | = | f | $$
but $$| f | |delta| ge | fastdelta | = | f | $$
which gives
$$ | fastdelta | = | f | $$
$$ |delta| ge | fastdelta | = 1 $$
With $ | delta | ge 1$, I believe I can derive a contradiction. I've tried integrating both sides, but that lead me nowhere. I'm not quite sure where to go from here. Any advice or tips?

real-analysis lebesgue-integral

share|cite|improve this question

asked Mar 24 at 20:36

NateNate

335

$endgroup$

add a comment | 

0

$begingroup$

I want to show there's no identity for the convolution of two functions. I've found posts that answer this for Riemann integrable functions, but I'm working with the convolution given by $fast g = int_{mathbb{R}^d} f(x-y)g(y) dy$, where $f, g$ are Lebesgue integrable. So, I want to show there does not exist a $deltain{L^1(mathbb{R}^d)}$ such that $fastdelta = f$.

My attempt:

Assume towards a contradiction that there is such a function $delta$. Then we have $fastdelta = f$
and hence,
$$ | fastdelta | = | f | $$
but $$| f | |delta| ge | fastdelta | = | f | $$
which gives
$$ | fastdelta | = | f | $$
$$ |delta| ge | fastdelta | = 1 $$
With $ | delta | ge 1$, I believe I can derive a contradiction. I've tried integrating both sides, but that lead me nowhere. I'm not quite sure where to go from here. Any advice or tips?

real-analysis lebesgue-integral

share|cite|improve this question

asked Mar 24 at 20:36

NateNate

335

$endgroup$

add a comment | 

$begingroup$

I want to show there's no identity for the convolution of two functions. I've found posts that answer this for Riemann integrable functions, but I'm working with the convolution given by $fast g = int_{mathbb{R}^d} f(x-y)g(y) dy$, where $f, g$ are Lebesgue integrable. So, I want to show there does not exist a $deltain{L^1(mathbb{R}^d)}$ such that $fastdelta = f$.

My attempt:

Assume towards a contradiction that there is such a function $delta$. Then we have $fastdelta = f$
and hence,
$$ | fastdelta | = | f | $$
but $$| f | |delta| ge | fastdelta | = | f | $$
which gives
$$ | fastdelta | = | f | $$
$$ |delta| ge | fastdelta | = 1 $$
With $ | delta | ge 1$, I believe I can derive a contradiction. I've tried integrating both sides, but that lead me nowhere. I'm not quite sure where to go from here. Any advice or tips?

real-analysis lebesgue-integral

share|cite|improve this question

asked Mar 24 at 20:36

NateNate

335

$endgroup$

share|cite|improve this question

asked Mar 24 at 20:36

NateNate

335

share|cite|improve this question

asked Mar 24 at 20:36

NateNate

335

asked Mar 24 at 20:36

NateNate

335

asked Mar 24 at 20:36

NateNate

335

1 Answer
1

active

oldest

votes

0

$begingroup$

Suppose there exists a function $delta in L^1_{loc}(mathbb{R}^d)$ such that $delta ast f = f$ for all $fin C_c^infty(mathbb{R}^d)$. Then let $phi_k$ be a sequence of smooth functions with $0leq phi_k leq 1$, $phi_k(0)=1$ and each with support contained in
$$
left[-frac{1}{k},frac{1}{k}right].
$$
The the dominated convergence theorem shows that
$$
int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x to 0.
$$
On the other hand,
$$
1 = phi(0) = deltaastphi(0) = int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x
$$
which is clearly a contradiction.

The above gives a slightly stronger result then what you asked for.

Alternatively, suppose for the sake of contradiction that there exists a function $delta in L^1(mathbb{R}^d)$ such that $fast delta = f$ for all $fin L^1(mathbb{R}^d)$.

Then, consider the simple $L^1$-function $rho = mathbb{1}_{S}$ where
$$
S= left[-frac{1}{2},frac{1}{2}right]^d
$$
For almost every $xin S$, we must have
$$
1 = rhoastdelta(x) = int_{mathbb{R}^d}rho(x-y) delta(y),mathrm{d}y
=int_{S+{x}} delta(y),mathrm{d}y.
$$
An application of the dominated convergence theorem therefore shows that
$$
int_{S} delta(y),mathrm{d}y = 1.
$$
Similarly, for every $x_0in mathbb{Z}^d$ we can derive
$$
int_{S_{x_0}} delta(y),mathrm{d}y = 1
$$
where
$$
S_{x_0}= left[x_0-frac{1}{2},x_0+frac{1}{2}right]^d.
$$
Finally, we conclude that
$$
int_{mathbb{R}^d} delta(y),mathrm{d}y
=sum_{x_0inmathbb{Z}^d}int_{S_{x_0}} delta(y),mathrm{d}y = infty
$$
which is clearly a contradiction.

share|cite|improve this answer

edited Mar 24 at 21:40

answered Mar 24 at 21:25

QuokaQuoka

1,330317

$endgroup$

add a comment | 

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0

$begingroup$

Suppose there exists a function $delta in L^1_{loc}(mathbb{R}^d)$ such that $delta ast f = f$ for all $fin C_c^infty(mathbb{R}^d)$. Then let $phi_k$ be a sequence of smooth functions with $0leq phi_k leq 1$, $phi_k(0)=1$ and each with support contained in
$$
left[-frac{1}{k},frac{1}{k}right].
$$
The the dominated convergence theorem shows that
$$
int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x to 0.
$$
On the other hand,
$$
1 = phi(0) = deltaastphi(0) = int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x
$$
which is clearly a contradiction.

The above gives a slightly stronger result then what you asked for.

Alternatively, suppose for the sake of contradiction that there exists a function $delta in L^1(mathbb{R}^d)$ such that $fast delta = f$ for all $fin L^1(mathbb{R}^d)$.

Then, consider the simple $L^1$-function $rho = mathbb{1}_{S}$ where
$$
S= left[-frac{1}{2},frac{1}{2}right]^d
$$
For almost every $xin S$, we must have
$$
1 = rhoastdelta(x) = int_{mathbb{R}^d}rho(x-y) delta(y),mathrm{d}y
=int_{S+{x}} delta(y),mathrm{d}y.
$$
An application of the dominated convergence theorem therefore shows that
$$
int_{S} delta(y),mathrm{d}y = 1.
$$
Similarly, for every $x_0in mathbb{Z}^d$ we can derive
$$
int_{S_{x_0}} delta(y),mathrm{d}y = 1
$$
where
$$
S_{x_0}= left[x_0-frac{1}{2},x_0+frac{1}{2}right]^d.
$$
Finally, we conclude that
$$
int_{mathbb{R}^d} delta(y),mathrm{d}y
=sum_{x_0inmathbb{Z}^d}int_{S_{x_0}} delta(y),mathrm{d}y = infty
$$
which is clearly a contradiction.

share|cite|improve this answer

edited Mar 24 at 21:40

answered Mar 24 at 21:25

QuokaQuoka

1,330317

$endgroup$

add a comment | 

0

$begingroup$

Suppose there exists a function $delta in L^1_{loc}(mathbb{R}^d)$ such that $delta ast f = f$ for all $fin C_c^infty(mathbb{R}^d)$. Then let $phi_k$ be a sequence of smooth functions with $0leq phi_k leq 1$, $phi_k(0)=1$ and each with support contained in
$$
left[-frac{1}{k},frac{1}{k}right].
$$
The the dominated convergence theorem shows that
$$
int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x to 0.
$$
On the other hand,
$$
1 = phi(0) = deltaastphi(0) = int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x
$$
which is clearly a contradiction.

The above gives a slightly stronger result then what you asked for.

Alternatively, suppose for the sake of contradiction that there exists a function $delta in L^1(mathbb{R}^d)$ such that $fast delta = f$ for all $fin L^1(mathbb{R}^d)$.

Then, consider the simple $L^1$-function $rho = mathbb{1}_{S}$ where
$$
S= left[-frac{1}{2},frac{1}{2}right]^d
$$
For almost every $xin S$, we must have
$$
1 = rhoastdelta(x) = int_{mathbb{R}^d}rho(x-y) delta(y),mathrm{d}y
=int_{S+{x}} delta(y),mathrm{d}y.
$$
An application of the dominated convergence theorem therefore shows that
$$
int_{S} delta(y),mathrm{d}y = 1.
$$
Similarly, for every $x_0in mathbb{Z}^d$ we can derive
$$
int_{S_{x_0}} delta(y),mathrm{d}y = 1
$$
where
$$
S_{x_0}= left[x_0-frac{1}{2},x_0+frac{1}{2}right]^d.
$$
Finally, we conclude that
$$
int_{mathbb{R}^d} delta(y),mathrm{d}y
=sum_{x_0inmathbb{Z}^d}int_{S_{x_0}} delta(y),mathrm{d}y = infty
$$
which is clearly a contradiction.

share|cite|improve this answer

edited Mar 24 at 21:40

answered Mar 24 at 21:25

QuokaQuoka

1,330317

$endgroup$

add a comment | 

$begingroup$

Suppose there exists a function $delta in L^1_{loc}(mathbb{R}^d)$ such that $delta ast f = f$ for all $fin C_c^infty(mathbb{R}^d)$. Then let $phi_k$ be a sequence of smooth functions with $0leq phi_k leq 1$, $phi_k(0)=1$ and each with support contained in
$$
left[-frac{1}{k},frac{1}{k}right].
$$
The the dominated convergence theorem shows that
$$
int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x to 0.
$$
On the other hand,
$$
1 = phi(0) = deltaastphi(0) = int_{mathbb{R}^d} delta(x)phi_k(x),mathrm{d}x
$$
which is clearly a contradiction.

The above gives a slightly stronger result then what you asked for.

Alternatively, suppose for the sake of contradiction that there exists a function $delta in L^1(mathbb{R}^d)$ such that $fast delta = f$ for all $fin L^1(mathbb{R}^d)$.

Then, consider the simple $L^1$-function $rho = mathbb{1}_{S}$ where
$$
S= left[-frac{1}{2},frac{1}{2}right]^d
$$
For almost every $xin S$, we must have
$$
1 = rhoastdelta(x) = int_{mathbb{R}^d}rho(x-y) delta(y),mathrm{d}y
=int_{S+{x}} delta(y),mathrm{d}y.
$$
An application of the dominated convergence theorem therefore shows that
$$
int_{S} delta(y),mathrm{d}y = 1.
$$
Similarly, for every $x_0in mathbb{Z}^d$ we can derive
$$
int_{S_{x_0}} delta(y),mathrm{d}y = 1
$$
where
$$
S_{x_0}= left[x_0-frac{1}{2},x_0+frac{1}{2}right]^d.
$$
Finally, we conclude that
$$
int_{mathbb{R}^d} delta(y),mathrm{d}y
=sum_{x_0inmathbb{Z}^d}int_{S_{x_0}} delta(y),mathrm{d}y = infty
$$
which is clearly a contradiction.

share|cite|improve this answer

edited Mar 24 at 21:40

answered Mar 24 at 21:25

QuokaQuoka

1,330317

$endgroup$

share|cite|improve this answer

edited Mar 24 at 21:40

answered Mar 24 at 21:25

QuokaQuoka

1,330317

answered Mar 24 at 21:25

QuokaQuoka

1,330317

answered Mar 24 at 21:25

QuokaQuoka

1,330317