(Fixed spelling errors on original post)
With both uniform and pointwise convergence we are dealing with a sequence of functions converging to a function. Don’t let the “pointwise” word confuse you. The “points” in this case are the x’s being substituted into the functions.
(Note I use x**n to mean x raised to the nth power and fn to represent the f(subscript n) functions.)
To prove pointwise convergence the value of x is fixed, for uniform convergence the value of x can take on any value in the space x is defined in. So from this result it’s easy to see if a sequence of functions is uniform convergent then it is also pointwise convergent, but not necessarily the converse. That is, a sequence may be pointwise convergent BUT there may be a point (one or more distinct x’s) where it is not uniform convergent.
For example, a sequence of functions defined as fn(x) = xn on the interval where 0 <= x <= 1 is pointwise convergent but NOT uniform convergent on the closed interval [0,1]. fn(x) converges to F(x) = 0 for 0 <= x < 1 and F(x) = 1 for x = 1 (note that fn(x) pointwise converges to 0 for x < 1 because the sequence xn converges to 0 for x < 1). To show fn(x) is NOT uniform convergent on the closed interval [0,1] we have to prove that the convergence of fn(x) depends not only on how close (usually called epsilon or e) we want to get to the limit but also on the value of x. By doing this we will contradict the definition of uniform convergence thereby prove that this sequence of functions is NOT uniform convergent on the interval [0,1].
So here is the meat of what you are after, what we have to do is work with the definition of the function sequence fn(x) along with the definition of uniform convergence. If we can prove fn(x) is not uniformly convergent on some sub-interval of [0,1], then it follows that fn(x) is not uniformly convergent on the entire interval [0,1]. Let’s look at the values of fn(x) when x=0 or x=1, the function is constant, so fn(x) is indeed uniform convergent at those points, but we are interested in the entire interval [0,1]. Now let’s consider the rest of the x-space, namely 0 < x < 1, in this case fn(x) = xn and we know xn converges to 0 in this interval because the sequence xn converges to 0 in this interval. So in this interval, fn(x) converges to F(x) which is the constant function 0, therefore |fn(x) - F(x)| = |fn(x) - 0| = xn for all n and for every x in this interval. But looking at the definition for uniform convergence we must be able to find a value for n where xn < e for each e we choose, so we need to solve for n. xn < e implies: n log x < log e implies: n > log e / log x. Since the value of n depends on the value of x, it contradicts the definition of uniform convergence and therefore, fn(x) is NOT uniformly convergent on the interval 0 < x < 1, which implies fn(x) is NOT uniformly convergent on the interval [0,1].
OK, now let’s do your problem.
fn(x) = (1-x)nx**n on the interval [0,1] (I believe that is your original function if I understood the notation correctly). First lets tackle pointwise convergence. Let’s figure out how this sequence behaves on this interval. Let’s write the sequence out in a little different format, you will agree that the original function can be defined as
fn(x) = nxn - nx(n+1).
Then we go to our trusty math reference to find out if this sequence is convergent on the interval [0,1], and if it is, what it converges to. Ok, believe me, this sequence converges to 0 on the interval [0,1], so the F(x) we were looking for is the constant value function F(x) = 0 (proof left to the reader … hehe). So, now we need to apply the conditions of pointwise convergence and see if this sequence of functions satisfies the conditions. What we need to do then is solve this equation for n: |fn(x) - F(x)| < e. Since the values of fn(x) are always > 0 on the interval [0,1] and F(x) = 0, the problem reduces to solving fn(x) < e for n. Substituting in the original equation for fn(x) we get
(1-x)nx**n < e
(Note I am a bit rusty with logs, check my work).
Or we can re-write the equation as n log (1-x)nx < log e
then finally
n > (log e) / (log (1-x)nx).
So, using the definition of pointwise convergence we must be able to say
for each x in the interval [0,1] and any e > 0, there exists an N such that when n > N, then |fn(x) - F(x)| < e.
If we set n to a value where n > (log e) / (log (1-x)nx) we have met the conditions necessary for pointwise convergence.
(Not sure I have completely solved the equation above, still need to solve for n, left as an exercise for the reader).
Now it’s time to prove this function is not uniform convergent on the interval [0,1]. After proving this function is pointwise convergent on [0,1] we realize that to find an n that satisfies the properties of uniform convergence we must also have a fixed value for x, and that contradicts the definition of uniform convergence, therefor this series of functions is not uniform convergent on the interval [0,1].
