mun wrote:

Shadow wrote:

alstat wrote:

Shadow wrote:

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

Of course, let the domain be

and let the function be

.

Since the definition of an even function is that

, can you have an even function whose domain is on only one side of zero?

The definition is

such that

, it just so happens that my set of

such that

is also in the domain is empty, so since we have a universal quantifier applied to an empty set, it is vacuously true that my function is even.

Good evening sir

sorry sir but one of my great sir has given this explanation (which confused me more )as he disagree with your explanation .His explanation is following

Any of the common definitions for even functions

A function such that

for all

in the domain of

; or

A function having its graph symmetrical with respect to the

axis;

in fact implies that the domain

of an even function

is itself symmetrical with respect to the origin, i.e.

. This is for example how the notion of even function is introduced in Romanian textbooks: a function

where

with

, and

for all

.

Thus speaking about an even function defined on

is disallowed, and not vacuously true (only because

does not exist).

I've never seen that assumption made, before. If that is your definition of an even function, then you may ignore what I've said, but I don't see any reason why one should assume the domain is symmetric and I never have. It's easy to *extend* even functions to a maximal, symmetric domain, but there's no reason do do that ahead of time IMO.

In that case you need to see if weakly decreasing is possible, if so, then the identically zero function works, otherwise the answer is no for the reason explained above (I'm assuming the function is

because this is a calculus question, you can get it for continuous functions without too much work as well.