4 ms·

The derivative is a linear operator, but it's not a bounded operator. That is, for example, the vector norm of f(x) = k·sin(x/k) → 0 when k→0, but the norm of d

by sampo 8y ago

The derivative is a linear operator, but it's not a bounded operator. That is, for example, the vector norm of f(x) = k·sin(x/k) → 0 when k→0, but the norm of d/dx f(x) does not. This also means that it's not continuous. Of the mappings between vector spaces, the most well behaving are the bounded linear operators, and the derivative doesn't belong to these. But yes, it's linear. Edit: Originally wrote f(x) = k·sin(k·x), but meant f(x) = k·sin(x/k).

giomasce 8y ago
You cannot speak of boundedness of operators if you don't specify the domain and codomain spaces. The derivation operator is for example bounded between C^1(R) and C^0(R), or between C^\infty(R) and C^\infty(R). Your example correctly proves that it is not bounded between C^0(R) and C^0(R).
hgibbs 8y ago
Depends on what space you define the derivative on. It is of course a bounded (and therefore continuous) operator from C^k to C^{k-1} for any positive integer K. Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces (as you need to make sense of what it means to be bounded), of which the most commonly dealt with are Banach spaces.
- enriquto 8y ago
  > Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces Reading your comment, I wondered how can you define bounded sets in a topological vector space (where you don't have a norm). The definition is cute: a set X if bounded if any neighborhood of 0 can be inflated to include the whole of X.
- adament 8y ago
  It also depends on what topology you put on your space, while the derivative operator is defined on all of C^k, that does not make it continuous. In fact as the GPs example shows, the topology you put on C^k and C^{k-1} must be so that uniform convergence does not imply convergence in C^k (which differs from many peoples intuitive notion of convergence of functions) or cos(x/k) converges to 0 in C^{k-1} (which is just plain weird).
soVeryTired 8y ago
Anyone have a concrete example of an open subset of C^k whose preimage under differentiation is not open? (Also, what's the most natural norm on C^k? The sup norm? I haven't done any functional analysis in years)
- hansen 8y ago
  If you want Cᵏ to be a normed space you do |f| = ∑ᵢ₌₀ᵏ sup|f⁽ⁱ⁾|. On open domains you can also use the topology that forces uniform convergence on all compact subsets. But this will only give you a metric space, no Banach space (but you’ll include unbouded functions). This is needed for studying Brownian motions with an unbounded time domain. Regarding the other question: If you take Cᵏ⁻¹ as a the co-domain differentiation will be continuous. IIRC to get unbounded linear maps defined on the whole Banach space you need the axiom of choice, you won’t be able to write one down. The problem with differential operators is that they are usually only defined on a dense subset of the domain, and there they are not bounded. E.g. in quantum mechanics the space of states is L² but all the interesting observables are differential operators. You can weasel out of this situation by defining them on a subset of “physical states” (e.g. smooth wave functions of rapid decay). But they aren’t continuous anymore (the spectrum is unbounded). But on Hilbert spaces everything mostly works out fine. Physicist usually ignore those technical problems and still don’t make mistakes.
msie 8y ago
So what level of mathematics is needed to understand the following thread? Undergrad real analysis? Grad math courses?
- minhazm423 8y ago
  hoping for an answer to this too
- sampo 8y ago
  > So what level of mathematics is needed to understand the following thread? First course in functional analysis. Before that, people usually take linear algebra, all the calculus courses, some kind of theorem-proving course for calculus (either analysis or real analysis or really rigorous versions of the calculus courses) and measure theory, maybe also topology.

by sampo 8y ago

giomasce 8y ago

You cannot speak of boundedness of operators if you don't specify the domain and codomain spaces. The derivation operator is for example bounded between C^1(R) and C^0(R), or between C^\infty(R) and C^\infty(R). Your example correctly proves that it is not bounded between C^0(R) and C^0(R).

hgibbs 8y ago

Depends on what space you define the derivative on. It is of course a bounded (and therefore continuous) operator from C^k to C^{k-1} for any positive integer K. Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces (as you need to make sense of what it means to be bounded), of which the most commonly dealt with are Banach spaces.

enriquto 8y ago
> Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces Reading your comment, I wondered how can you define bounded sets in a topological vector space (where you don't have a norm). The definition is cute: a set X if bounded if any neighborhood of 0 can be inflated to include the whole of X.
adament 8y ago
It also depends on what topology you put on your space, while the derivative operator is defined on all of C^k, that does not make it continuous. In fact as the GPs example shows, the topology you put on C^k and C^{k-1} must be so that uniform convergence does not imply convergence in C^k (which differs from many peoples intuitive notion of convergence of functions) or cos(x/k) converges to 0 in C^{k-1} (which is just plain weird).

soVeryTired 8y ago

Anyone have a concrete example of an open subset of C^k whose preimage under differentiation is not open? (Also, what's the most natural norm on C^k? The sup norm? I haven't done any functional analysis in years)

hansen 8y ago
If you want Cᵏ to be a normed space you do |f| = ∑ᵢ₌₀ᵏ sup|f⁽ⁱ⁾|. On open domains you can also use the topology that forces uniform convergence on all compact subsets. But this will only give you a metric space, no Banach space (but you’ll include unbouded functions). This is needed for studying Brownian motions with an unbounded time domain. Regarding the other question: If you take Cᵏ⁻¹ as a the co-domain differentiation will be continuous. IIRC to get unbounded linear maps defined on the whole Banach space you need the axiom of choice, you won’t be able to write one down. The problem with differential operators is that they are usually only defined on a dense subset of the domain, and there they are not bounded. E.g. in quantum mechanics the space of states is L² but all the interesting observables are differential operators. You can weasel out of this situation by defining them on a subset of “physical states” (e.g. smooth wave functions of rapid decay). But they aren’t continuous anymore (the spectrum is unbounded). But on Hilbert spaces everything mostly works out fine. Physicist usually ignore those technical problems and still don’t make mistakes.

msie 8y ago

So what level of mathematics is needed to understand the following thread? Undergrad real analysis? Grad math courses?

minhazm423 8y ago
hoping for an answer to this too
sampo 8y ago
> So what level of mathematics is needed to understand the following thread? First course in functional analysis. Before that, people usually take linear algebra, all the calculus courses, some kind of theorem-proving course for calculus (either analysis or real analysis or really rigorous versions of the calculus courses) and measure theory, maybe also topology.