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This is a past exam question which im not sure how to solve..

${For \ n=1,2,3,... consider \ f_n: [0,2] \rightarrow \mathbb{R} \ given \ by}$

$f_n(x)=$ $\left\{\begin{array}{l l}nx & \quad \text{0 $\leq x <\dfrac{1}{n} $ }\\2-nx & \quad\text{$\dfrac{1}{n} \leq x < \dfrac {2}{n}$} \\0 & \quad\text{otherwise} \end{array} \right.$

$a)Determine \ the \ pointwise \ limit \ f(x)=\displaystyle\lim_{n\to \infty}f_n(x)$

$b)Does \ f_n \rightarrow f \ with \ respect \ to \ the \ 1-norm ||g||_1 = \displaystyle\int^2_0 |g(x)|\ dx?$

$c)Does f_n \rightarrow f \ uniformly?$

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Try sketching the graphs of the first few $f_n$ and see what you can surmise. – David Mitra Feb 21 '13 at 20:01
@DavidMitra I have drawn the graph for the few few $f_n$ but im not sure how this helps. – Mathsstudent147 Feb 21 '13 at 20:32
Can you see that the pointwise limit $0$, to begin with? – 1015 Feb 21 '13 at 20:33
Look at them. Note the graph of $f_n$ is $0$ most everywhere, except for a spike of height $1$ centered at $x=1/n$. As $n$ grows larger, the spike "heads towards" the $y$-axis. So what is the pointwise limit? For b), what do the areas under the graphs tend to? For c), is it true that for large $n$, $f_n$ is uniformly close to the pointwise limit? – David Mitra Feb 21 '13 at 20:35
@DavidMitra The areas under the graph tend to 0 c) For large n, $f_n$ is uniformly close to the pointwise limit, which is 0. – Mathsstudent147 Feb 21 '13 at 20:44

At a) $f_n(0)=0$ for every $n$. For $x>0$ there exists $n_0$ such that $\frac{1}{n}<\frac{1}{n_0}<x$ for every $n>n_0$. Consequently $f_n(x)\to 0$ for every $x\in [0,2]$.

At b) $f_n$ is a triangle with height $1$ and width $2/n$. Consequently $||f_n||=1/(4n)\to 0=||f||$

At c) no since $\max\{f_n(x)-f(x)\}=1$ which is arbitrary large

Thanks for the remarks :) was thinking of a different funtion somehow

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I still don't really understand part b.... and don't the triangles have height n? – Mathsstudent147 Feb 21 '13 at 20:41
For c), the $\max$ is $1$. – David Mitra Feb 21 '13 at 20:48
thanks again :) the maximum of the tirangles is in x=1/n and there the value is 1. By the way, the function is continuous. – Quickbeam2k1 Feb 21 '13 at 20:49

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