Sequences and continuous functions

[the_kid]

Homework Statement

a) Let {s_{n}} and {t_{n}} be two sequences converging to s and t. Suppose that s_{n} < (1+\frac{1}{n})t_{n}

Show that s \leqt.

b) Let f, g be continuous functions in the interval [a, b]. If f(x)>g(x) for all x\in[a, b], then show that there exists a positive real z>1 such that f(x)\geqzg(x) for all x\in[a, b].

Homework Equations

The Attempt at a Solution

Ok, so I've already done part a. I'm trying to figure out part b. I think my ideas are on the right track, but I'm looking for some help fleshing them out a bit more.

Argue by contradiction. Suppose there does not exist z such that f(x)\geqzg(x). I'm not sure exactly what this implies in terms of deriving a contradiction; some general guidance would be appreciated it. Also, I'm using the definition that the notion that f(x) is continuous at c is equivalent to the following:

For every sequence {s_{n}} in the domain of f converging to c, one has

lim, n-->\infty f(s_{n})=f(c). I somehow want to construct an analogous argument to that in part a, but I'm not sure how the z is similar to the (1+\frac{1}){n}) term. Can anyone help with this?

 

[dirk_mec1]

I think this can go easier, define h:

f(x)-g(x)=h(x)

because f,g are continuous in [a,b], h is continuous in [a,b] but then h(x) has a minimum >0 in [a,b] call it c thus:

f \geq g+c = g \left( 1+\frac{c}{g(x)} \right)

This holds for all x in [a,b] thus:

f \geq g+c = g \left( 1+\frac{c}{ ||g(x)||_{\infty} } \right) = zg(x)with ||g(x)||_{\infty} =sup_{x \in [a,b]} g(x) and z= 1+\frac{c}{ ||g(x)||_{\infty} }Something like this should work ( I think the part with the inf term should be replaced by something else).