# Answer to Question #16429 in Functional Analysis for Ciaran Duffy

Question #16429

Suppose the function f is continuous on the interval [a, b] and never zero on [a, b].

Is it possible that f(z) < 0 for some z ∈ [a, b] and f(w) > 0 for some w ∈ [a, b]?

Explain your answer.

Is it possible that f(z) < 0 for some z ∈ [a, b] and f(w) > 0 for some w ∈ [a, b]?

Explain your answer.

Expert's answer

This is not impossible due the intermediate value theorem.

This theorem

claims that if a continuous function

f : [z,w] --> R

on the interval

[z,w] is such that

f(z) and f(w) are non-zero and have distinct signs,

e.g.

either

f(z)>0 and f(w)<0

or

f(z)<0 and

f(w)>0,

then there exists a point c in [z,w] such that f(c)=0.

In

our case suppose

f(z)<0 for some z ∈ [a, b] and f(w)>0 for some w ∈

[a, b]

We can assume that z<w.

Then by the above theorem ther eshould

exists a point c in [z,w] such that f(c)=0.

But c also belongs to [a,b],

and so f is not never zero on [a,b].

This gives a contradiction, and so

either

f>0

or

f<0

of all of [a,b].

This theorem

claims that if a continuous function

f : [z,w] --> R

on the interval

[z,w] is such that

f(z) and f(w) are non-zero and have distinct signs,

e.g.

either

f(z)>0 and f(w)<0

or

f(z)<0 and

f(w)>0,

then there exists a point c in [z,w] such that f(c)=0.

In

our case suppose

f(z)<0 for some z ∈ [a, b] and f(w)>0 for some w ∈

[a, b]

We can assume that z<w.

Then by the above theorem ther eshould

exists a point c in [z,w] such that f(c)=0.

But c also belongs to [a,b],

and so f is not never zero on [a,b].

This gives a contradiction, and so

either

f>0

or

f<0

of all of [a,b].

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