Uniform Convergence of Fn(x)=nx(1-x^2)^n on [0,1]?

[math-chick_41]

does Fn(x)= nx(1 - x^2)^n converge uniformly on [0,1]?
my first instinct was yes it converges uniformly to 0
but I can't seem to show that using the definition.

i get |nx(1 - x^2)^n|<=|nx|<=n for x in [0,1]

any tip or hint would be helpful
thanks

 

[matt grime]

What is the sup norm of f_n? That should give you the answer. Hint, it might help to remember basic analysis like turning points and points of infelxion...

 

[math-chick_41]

for n large and x really small, namely x=1/n
i get that the sup{|f_n(x)|:x in [0,1]} = 1
so then it doesn't converge uniformly.
is that right?

 

[matt grime]

That'll be right as long as you believe that the limit, if it were to exist, were the zero function, or something like it. After all, you need to have the notional idea of what it might converge to if it were to converge (and then to show that it is not uniform convergence). So, show it converges pointwise, and then show it is not uniform (if it were uniform then it would certainly be pointwise).

I don't agree that is important we say *for n really large* and *x really small* the the sup is something, rather than for ANY n the sup is..., after all the sup norm ought to be at least something independent of the n.