Let f satisfy f(x + y) = f(x) + f(y), f continuous at x = 0.

[marlen19861]

Suppose f satisfies the relation f(x+y)=f(x)+f(y) and f continuous at 0. Show that f is continuous at all points.

Any help? Thank you

 

[pka]

Suppose f satisfies the relation f(x+y)=f(x)+f(y) and f continuous at 0. Show that f is continuous at all points.

Can you prove that \(\displaystyle f(0)=0\)? Hint: \(\displaystyle f(x)=f(x+0)\).
From continuity at zero \(\displaystyle \varepsilon > 0\; \Rightarrow \;\left( {\exists \delta > 0} \right)\left[ {\left| x \right| < \delta \; \Rightarrow \;\left| {f(x)} \right| < \varepsilon } \right]\)
From which it follows that \(\displaystyle \left( {\forall y} \right)\;\& \;\lambda \in \left( { - \delta ,\delta } \right)\; \Rightarrow \;\left| {f(y + \lambda ) - f(y)} \right| < ?\)
Can you do the proof now?

 

[mmm4444bot]

I cannot help with this posted exercise, but the relationship interests me.

Off the top of my head, I can think of three definitions for function f, such that the sum f(a) + f(b) equals f(a+b) for arbitrary real numbers a and b.

f(x) = 0

f(x) = x

f(x) = -x

Now, the top of my head is empty. Hmmm.

I've made the assumption that there must be more, since these three are clearly continuous at the origin.

I suspect that I'll need to ponder quite a while to come up with another function (without considering "unusual" things, like modular arithmetic) that satisfies this relationship.

A brain teaser ...

~ Mark

 

[daon]

Mark, there are infinitely many :wink:

If r is a real number and f(x)=x, then g_r(x)=r*f(x) is just uncountably many examples.

Look up homomorphism.

 

[arman_phsa]

I don't know exactly if my answer is correct or not but i solved the problem in this way:
we know that f(0)=0 { f(0)=2f(0) }
so f(0)=f(x-x)=f(x)+f(-x)=0 so f(-x)=-f(x) *
now we know that f is continues at 0 so we will use the definition :
for all e>0 (epsilon) we have d>0 (delta) that [ |x-0|<d |f(x)-f(0)|<e ] so [ |x|<d , |f(x)|<e ]
we know the above expression is correct now we will pick "a" whether we want and set x=X-a and use in above :
|X-a|<d , |f(X-a)|<e so |f(X)+f(-a)|<e and from "*" we have :
|X-a|<d , |f(X)-f(a)|<e and this is the continues definition for x { lim(x goto a)f(x)=f(a) } and because we had no limit in choosing "a" the function is continues in every point of its domain
I don't know if my answer is right or not so please send your comments about this answer TY