Continuity question help

SWISH99

Suppose g(-x) = -g(x) for any x. Show that if g is continuous at 0 then g(0)=0

(edited 6 years ago)

Reply 1

_gcx

Study Forum Helper

Original post by SWISH99

Suppose g(-x) = -g(x) for any x. Show that if g is continuous at 0 then g(0)=0

Set

x=0

Reply 2

NotNotBatman

Original post by SWISH99

Suppose g(-x) = -g(x) for any x. Show that if g is continuous at 0 then g(0)=0

In this case g(x) is an odd function, so g(-0) = -g(0). Then it's an easy implication from there.

Reply 3

atsruser

Original post by NotNotBatman

In this case g(x) is an odd function, so g(-0) = -g(0). Then it's an easy implication from there.

You haven't mentioned continuity though.

Reply 4

DFranklin

Original post by atsruser

You haven't mentioned continuity though.

Since the result holds without needing continuity, this is perhaps unsurprising.

(Possibly the original question only specified f(x)=-f(-x) for x not equal to 0; the conclusion then requires continuity).

Reply 5

atsruser

Original post by DFranklin

y=1/x is a counterexample, no? Seems to me that continuity guarantees definedness at x=0.

Reply 6

_gcx

Study Forum Helper

Original post by atsruser

y=1/x is a counterexample, no? Seems to me that continuity guarantees definedness at x=0.

I agree, The question is probably looking for you to justify letting

x=0

with the fact that

g

is continuous at 0.

Reply 7

NotNotBatman

Original post by atsruser

You haven't mentioned continuity though.

But the question assumes continuity, so g(x) is defined at x=0.

Reply 8

DFranklin

Original post by atsruser

y=1/x is a counterexample, no?

I wouldn't say it's a counterexample to the question as quoted. Unless told otherwise, there's no reason to assume that 0 is a "special" point as far as being able to apply f(x) = -f(-x), or in terms of f being defined (in fact, there are strong reasons to say it makes no sense at all to suggest f might not be defined at 0).

As I said in my reply, I suspect there's some missing preamble to the question.

Seems to me that continuity guarantees definedness at x=0.

Well yes, but note that continuity guarantees from the definition that both

f(0) is defined

and

\lim_{x \to 0} f(x) = f(0)

In other words, there's no "analysis" required to deduce f(0) is defined - it's directly one of the conditions of continuity.

The only way I see any "meat" here is if as I said before, the question did not mean to state that f(x) = -f(-x) holds at x = 0.

Then if you assume

f(0) = \epsilon \neq 0

, by using the definition of continuity and considering points slightly above and slightly below 0, you can deduce a contradiction.

Reply 9

atsruser

Original post by DFranklin

Sorry - short of time. But it seems that we have to interpret the question wording rather widely, else the mention of continuity seems to be irrelevant and the question pretty vacuous - if we assume g(x) is defined at 0, then it's trivial.

Reply 10

atsruser

Original post by NotNotBatman

But the question assumes continuity, so g(x) is defined at x=0.

I think the question tries to refer awkwardly to two different g's - the first lot are odd, and the second lot are continuous. It's only the intersection of both where the result holds. Anyway, gotta go - I'll leave you to further learned discussions.

Reply 11

DFranklin

Original post by atsruser

But deducing g(x) is defined at 0 from g cts at 0 is also trivial. It's literally part of the definition of continuity.

The OP posted the full question elsewhere; there's not actually any more detail, but from the extra context, I'm fairly sure the lecturer was *aiming* at establishing something along the lines that "if f is defined for x not 0, and f (where defined) is odd , then the only way to define f(0) such that f is cts is f(0)=0".

Unfortunately the actual question asked doesn't say anything about "x not 0", which makes the whole thing meaningless.