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 $\leq$t.

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)$\geq$zg(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)$\geq$zg(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).