Question about Linear systems

robert bristow-johnson wrote:
...
> it can be shown that if the additivity
> property holds (for just two added inputs), what i usually like to
> call "superposition":
>
>      T{ x1(t) + x2(t) } = T{ x1(t) } + T{ x2(t) }
>
>      for *any* given x1(t) and x2(t)
>
> if that property holds, you can easily show that the scaling property
> holds
>
>      T{k*x } = k*T{x}
>
> for any *rational* k.  it's a little harder to do for the irrational
> k, which requires an additional assumption about the continuous nature
> of the system T{..},

As I mentioned already, continuity of T is sufficient but not
necessary for the scaling property. We all know that d/dx (a f(x)) = a
d/dx f(x) (the homogeneity property holds), yet the differential
operator is not continuous. As it turns out, it is not much harder to
show that if

T{q x}

holds for all rational numbers q, it also holds for all irrational
numbers r.

To see this, write

r = lim_{k -> oo} q_k,

where {q_k} is a sequence of rational numbers (every irrational number
can be written as a sequence of rational numbers, because the rational
numbers are dense in the real numbers). Now let

T:X->Y

be an additive operator, and X and Y two complete, normed vector
spaces. Furthermore, T is homogeneous for rational numbers, ie.

T{q x} = q T{x}

for all x in X and for all rational numbers q. For x in X, denote

y = T{x}

in Y. Firstly q_k x -> r x (for k->oo), because

|| q_k x - r x || = |q_k - r| ||x|| -> 0 for k->oo. Similarly

q_k y -> q y. So we have

lim_{k -> oo} T{q_k x} = T{ r x}
lim_{k -> oo} q_k T{x} = r T{x}.

But T{q_k x} = q_k T{x} for all k (ie. the two sequences are equal),
therefore

r T{x} = T{r x}

(the limits must be equal also).

Note that for this simple proof I have never used the continuity of T,
only the homogeneity for rational numbers and basic properties of
Banach spaces (homogeneity of the norm, completeness and uniqueness of
limits of Cauchy sequences).

And this begs the question: if additivity implies linearity, why are
we always talking about linear operators, if the more parsimonious
definition of additive operators would suffice?

Regards,
Andor

Andor wrote:
> robert bristow-johnson wrote:
> ...
>> it can be shown that if the additivity
>> property holds (for just two added inputs), what i usually like to
>> call "superposition":
>>
>>      T{ x1(t) + x2(t) } = T{ x1(t) } + T{ x2(t) }
>>
>>      for *any* given x1(t) and x2(t)
>>
>> if that property holds, you can easily show that the scaling property
>> holds
>>
>>      T{k*x } = k*T{x}
>>
>> for any *rational* k.  it's a little harder to do for the irrational
>> k, which requires an additional assumption about the continuous nature
>> of the system T{..},
> 
> As I mentioned already, continuity of T is sufficient but not
> necessary for the scaling property. We all know that d/dx (a f(x)) = a
> d/dx f(x) (the homogeneity property holds), yet the differential
> operator is not continuous. As it turns out, it is not much harder to
> show that if
> 
> T{q x}
> 
> holds for all rational numbers q, it also holds for all irrational
> numbers r.
> 

I believe you're mistaken.

> To see this, write
> 
> r = lim_{k -> oo} q_k,
> 
> where {q_k} is a sequence of rational numbers (every irrational number
> can be written as a sequence of rational numbers, because the rational
> numbers are dense in the real numbers). Now let
> 
> T:X->Y
> 
> be an additive operator, and X and Y two complete, normed vector
> spaces. Furthermore, T is homogeneous for rational numbers, ie.
> 
> T{q x} = q T{x}
> 
> for all x in X and for all rational numbers q. For x in X, denote
> 
> y = T{x}
> 
> in Y. Firstly q_k x -> r x (for k->oo), because
> 
> || q_k x - r x || = |q_k - r| ||x|| -> 0 for k->oo. Similarly
> 
> q_k y -> q y. So we have

I assume q = r (the limit of the q_k).

> 
> lim_{k -> oo} T{q_k x} = T{ r x}

But this doesn't follow, does it?  All you have established is that

T(lim_k (q_k x)) = T((lim_k q_k) x) = T(rx).

To get from here to your assertion (moving the limit outside the
operator T) requires continuity of T.

> lim_{k -> oo} q_k T{x} = r T{x}.
> 
> But T{q_k x} = q_k T{x} for all k (ie. the two sequences are equal),
> therefore
> 
> r T{x} = T{r x}
> 
> (the limits must be equal also).
> 
> Note that for this simple proof I have never used the continuity of T,
> only the homogeneity for rational numbers and basic properties of
> Banach spaces (homogeneity of the norm, completeness and uniqueness of
> limits of Cauchy sequences).
> 
> And this begs the question: if additivity implies linearity, why are
> we always talking about linear operators, if the more parsimonious
> definition of additive operators would suffice?
> 
> Regards,
> Andor
> 

An additive but not continuous operator need not be linear.  For
example, the reals R form an infinite-dimensional vector space over the
rationals Q, and one can choose a basis B (necessarily infinite) for R
over Q containing 1 and your favorite irrational number, e.g. pi.  Any
function f : B -> R has a unique extension to a Q-linear function
f':B -> R; choose f so that f(1) = f(pi) = 1.  Then f', being Q-linear,
 is additive but f'(pi 1) = f'(pi) = 1 =/= pi f'(1).  Call this example
pathological if you like but it does show that additivity without
continuity is weaker than linearity.

Hope that helps,

Robert E. Beaudoin

Robert E. Beaudoin wrote:
> Andor wrote:
> > robert bristow-johnson wrote:
> > ...
> >> it can be shown that if the additivity
> >> property holds (for just two added inputs), what i usually like to
> >> call "superposition":
>
> >>      T{ x1(t) + x2(t) } = T{ x1(t) } + T{ x2(t) }
>
> >>      for *any* given x1(t) and x2(t)
>
> >> if that property holds, you can easily show that the scaling property
> >> holds
>
> >>      T{k*x } = k*T{x}
>
> >> for any *rational* k.  it's a little harder to do for the irrational
> >> k, which requires an additional assumption about the continuous nature
> >> of the system T{..},
>
> > As I mentioned already, continuity of T is sufficient but not
> > necessary for the scaling property. We all know that d/dx (a f(x)) = a
> > d/dx f(x) (the homogeneity property holds), yet the differential
> > operator is not continuous. As it turns out, it is not much harder to
> > show that if
>
> > T{q x}
>
> > holds for all rational numbers q, it also holds for all irrational
> > numbers r.
>
> I believe you're mistaken.
>
>
>
>
>
> > To see this, write
>
> > r = lim_{k -> oo} q_k,
>
> > where {q_k} is a sequence of rational numbers (every irrational number
> > can be written as a sequence of rational numbers, because the rational
> > numbers are dense in the real numbers). Now let
>
> > T:X->Y
>
> > be an additive operator, and X and Y two complete, normed vector
> > spaces. Furthermore, T is homogeneous for rational numbers, ie.
>
> > T{q x} = q T{x}
>
> > for all x in X and for all rational numbers q. For x in X, denote
>
> > y = T{x}
>
> > in Y. Firstly q_k x -> r x (for k->oo), because
>
> > || q_k x - r x || = |q_k - r| ||x|| -> 0 for k->oo. Similarly
>
> > q_k y -> q y. So we have
>
> I assume q = r (the limit of the q_k).

Yes.

>
>
>
> > lim_{k -> oo} T{q_k x} = T{ r x}
>
> But this doesn't follow, does it?  

That's true - it only follows directly if T is continuous. Damn.

> All you have established is that
>
> T(lim_k (q_k x)) = T((lim_k q_k) x) = T(rx).

So we have

r T(x) = (lim_k q_k) T(x)
= lim_k (q_k T(x))
= lim_k T(q_k x).
=?= T(lim_k (q_k x))

>
> To get from here to your assertion (moving the limit outside the
> operator T) requires continuity of T.

But I know that continuity is not necessary for the last question-
equation above (the differential operator is homogeneous but not
continuous). Is there no other way?

> An additive but not continuous operator need not be linear. For
> example, the reals R form an infinite-dimensional vector space over the
> rationals Q, and one can choose a basis B (necessarily infinite) for R
> over Q containing 1 and your favorite irrational number, e.g. pi.  Any
> function f : B -> R has a unique extension to a Q-linear function
> f':B -> R; choose f so that f(1) = f(pi) = 1.  Then f', being Q-linear,
>  is additive but f'(pi 1) = f'(pi) = 1 =/= pi f'(1).  Call this example
> pathological if you like but it does show that additivity without
> continuity is weaker than linearity.
>
> Hope that helps,
>
> Robert E. Beaudoin

Thanks for the input.

Regards,
Andor

Andor wrote:

[snip]

> 
> So we have
> 
> r T(x) = (lim_k q_k) T(x)
> = lim_k (q_k T(x))
> = lim_k T(q_k x).
> =?= T(lim_k (q_k x))
> 
>> To get from here to your assertion (moving the limit outside the
>> operator T) requires continuity of T.
> 
> But I know that continuity is not necessary for the last question-
> equation above (the differential operator is homogeneous but not
> continuous). Is there no other way?
> 

[snip]

OK, I misspoke slightly:  continuity of T is sufficient but not
necessary for the equality you decorated with a question mark.  Of
course what I meant to say was that this equality is not true for an
arbitrary T (and of course you see that now).  It's enough that T be
continuous on every one-dimensional subspace of its domain.

Robert E. Beaudoin

On Oct 14, 7:11 am, Andor <andor.bari...@gmail.com> wrote:
...
> And this begs the question: if additivity implies linearity, why are
> we always talking about linear operators, if the more parsimonious
> definition of additive operators would suffice?

maybe, because of the theoretical difficulty of extending
superposition (the additivity property) to scaling using irrational
scalers, or that it's just pedagogically easier to start with the two
axioms.  but, really, as far as i'm concerned, additivity implies
linearity, just as we can deal with dirac delta functions like
functions and just as i'm sufficiently satisfied with the model of
Riemann Integration (rather than restorting to Lebesgue).  you'ld have
to get very anal about it mathematically (to the extent that it is not
useful for engineering) to take issue with the practicality and safety
of that.

r b-j