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For $n in mathbbN$, define the formula, $$f_n(x)= fracx2n^2x^2+8,quad x in [0,1].$$ Prove that the sequence $f_n$ converges uniformly on $[0,1]$, as $n to infty$.

I know that the definition says $f_n$ converges uniformly to $f$ if given $forall epsilon gt 0$, $forall n geq N$, such that $|f_n(x) - f(x)| lt epsilon, forall n geq N$ and $forall x in [0,1].$

I looked first at the pointwise convergence and found that $$lim_n rightarrow infty fracx2n^2x^2+8 = 0, forall x in [0,1].$$

So how do I use this to choose an $n geq N$ such that $|f_n(x) - f(x)| lt epsilon$ ?

Right now, I have

"proof: Let $epsilon > 0, exists N in mathbbN$ such that, n $geq N Rightarrow frac12n^2+8 lt epsilon$,

by $|f_n(x) - 0| = |fracx2n^2x^2+8| leq |fracx^22n^2x^2+8| leq frac12n^2x^2+8 ;;;; forall x in [0,1].$

Since $lim_n rightarrow infty fracx2n^2x^2+8 = 0, forall x in [0,1]$, $f_n(x)$ will converge uniformly to $0$ on $[0,1]$."

Is this correct? Am I missing something? Is something not correct? I'm unsure about my choice of $N$. Please & thanks!

related problem: (I), (II), (III). Here is a systematic technique. In order to find $sup_0leq xleq 1 |f_n(x)-f(x)| $, you need to maximize the function $Big|fracx2n^2x^2+8Big|$ over the interval $[0,1]$. Now, let

$$ g(x)=fracx2n^2x^2+8 implies g'(x) = frac4-n^2 x^2(2n^2x^2+8)^2=0 implies x=frac2n$$

gives the max of the function $g(x)$ which is $g(2/n)=1/8n$. You can check this by checking the sign of $g''(x)$ which should be $< 0$. Hence we have

$$ sup_0leq xleq 1 |f_n(x)-f(x)|= sup_0leq xleq 1 Big|fracx2n^2x^2+8Big|=frac18n < epsilon. $$

You can write
$$f_n(x)=1over n g(n>x)qquad(0leq xleq 1)$$
with
$$g(t)=tover 2t^2+8qquad(0leq t<infty) .$$
Since $g(t)$ converges to $0$ for $ttoinfty$ we already can say that $g$ is bounded. A quantitative estimate can be obtained as follows: For $t>0$ one has
$$0<g(t)=1over 8 2overtover 2+2over tleq1over 8 .$$
Therefore we now have
$$|f_n(x)|leq1over 8nqquad(ngeq1, 0leq xleq1) ,$$
which shows that the $f_n$ converge uniformly to $0$ on $[0,1]$.

Another way to maximize :

$dfracx2n^2x^2+8leq dfracx8nx=dfrac18n$ where we used the AM-GM inequality
:http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means to show

$dfrac2n^2x^2+82 geq sqrt 2n^2x^2 cdot 8 $ Applying this easy inequality is often a good first way to find a supremum for the function if you have a summation involved.
So since the function converges pointwise to $f(x)=0$ We have the following result:

$lim_n to inftysup_0leq xleq 1 |f_n(x)-f(x)|=lim_n to infty|dfrac18n|=0$

Pointwise convergence is not enough to say that the function converges uniformly. Here, $$f_n(x)= fracx2n^2x^2+8,x in [0,1]$$ has pointwise convergence to $f(x)=0$, so by definition $$|f_n(x)-f(x)|=bigg|fracx2n^2x^2+8-0bigg|= bigg|fracx2n^2x^2+8bigg|< frac12n$$
This shows that that the function is not uniformly convergent.