f(x+y)=f(x)+f(y)=>f continuous??
Rocco
Here is a problem that i can't solve:
f : R->R (R are the reals) is a function such that f(x+y)=f(x)+f(y) for all
(x,y) in R^2.
Is f necessarily continous? -in fact, can we easily find a counterexample ?-
thanks for your help!!
Arturo Magidin
No, f is not necessarily continuous. No, we cannot find an easy
counterexample; in fact, it is not possible to write down one, if I'm
not mistaken.
The condition given implies that:
f(ax) = af(x)
for any rational number x.
Consider R as a vector space over Q. Then, using Zorn's Lemma, one can
find a basis for R as a vector space over Q (an infinite basis), say
(v_i), i in I.
Then every element of R may be written uniquely as
x = sum (a_i*v_i)
with a_i rational, and all but a finite number of them equal to zero.
Fix a specific element of the basis, say v_0. Let f be the "projection
onto the v_0-th coordinate". That is, let f(x) = a_0, where x = sum
(a_i*v_i).
Then f is a linear transformation, so in particular f(x+y)=f(x)+f(y)
for all x and all y in R.
Yet it is not hard to verify that f is not continuous.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
José Carlos Santos
No, f is not necessarily continous. Consider R as a vector space over Q
(the rationals). Take a basis B of that space and consider a linear
function f such that f(v) = 1 for some element v of B and that takes the
value 0 at every other element of B. Then f(x + y) = f(x) + f(y). But
if f : R -> R has this property and is also continuous, then, for every
x in R, f(x) = f(1).x, and so the function mentioned above cannot be
continuous.
Best regards
Jose Carlos Santos
David Kastrup
Well, f(0) is trivial, so is f(nx)=n f(x) and from there f((p/q)x) =
(p/q)f(x).
So fixing one rational apart from 0 fixes them all. Now you just
need a good scheme to go goofy on the irrationals, or at least a
subset of them...
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Karl Forsberg
As others have posted - the answer is no. However if you assume that f
is measurable (a fairly weak assumption) then the answer is yes.
Robert Israel
Karl Forsberg <MBPCLI...@spammotel.com> wrote:
>Rocco wrote:
>> f : R->R (R are the reals) is a function such that f(x+y)=f(x)+f(y) for all
>> (x,y) in R^2.
>> Is f necessarily continous? -in fact, can we easily find a counterexample ?-
>As others have posted - the answer is no. However if you assume that f
>is measurable (a fairly weak assumption) then the answer is yes.
In fact, it's a nice exercise to show that if f is measurable and
satisfies the equations f(2x) = 2 f(x) and f(3x) = 3 f(x), then f(x)/x
is constant for x > 0.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Robert Israel
Robert Israel <isr...@math.ubc.ca> wrote:
>In fact, it's a nice exercise to show that if f is measurable and
>satisfies the equations f(2x) = 2 f(x) and f(3x) = 3 f(x), then f(x)/x
>is constant for x > 0.
Oops: I meant then f(x)/x is constant almost everywhere for x > 0.
Marko
A very good discussion of Cauchy's functional equation can be found in
C.H.Morgan's B.A. Thesis on functional equations.
j.e.mebius
In my student's years in the 1960s I asked the same question to my teacher prof.
dr F. van der Blij. He told me that to construct a discontinuous linear function
on the reals, one needs a so-called Hamel basis of the reals as a vector space
over the field Q of the rationals.
With the help of the Axiom of Choice one proves the existence of such a basis.
Hamel bases are necessarily uncountable. As far as I know, an explicit
construction is unknown, and the problem of contructing a Hamel basis is
undecidable.
BTW, discontinuous linear functions on Q(sqrt(2)), and indeed on any real
algebraic number field (always finite-dimensional as a linear space over Q and
always not complete as a metric space with the usual distance function), are
easily constructed.
Hope this helps you to some extent.
Best regards: Johan E. Mebius
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