Convergence in $(L_P(a,b),\|.\|_p)$ doesn't imply convergence in $(C(a,b),\sup)$ 
Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),\|.\|_p)$? 
It seems to me for the considered function that $\|f_n(x)-0\|_p \leq \frac{1}{(1-p)} (\frac{(1+n)^{1-p}}{n} - \frac{1}{n}) < \epsilon$
then for all $n>N$, $f_n(x) \to 0$ uniformly in $(L_P(a,b),\|.\|_p)$ as $N$ depends on $\epsilon$ and $p$.
And If I'm wrong and it is pointwise convergent in $(L_P(a,b),\|.\|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),\sup)$?
 A: First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n \to f$ wrt. to $||\, .\, ||_\infty$ if and only if $f_n \to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.
$f_n \to f$ with respect to $||\, .\, ||_p$ means only that 
\begin{align}
\int_a^b |f_n(t) - f(t)|\, d\lambda(t) \to 0 \qquad \text{as} \qquad n \to \infty
\end{align} 
It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n \to f$ with respect to the $L^p$-norm, but $f_n(x) \to f(x)$ at no point $x \in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.
Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $\forall x \in [a,b]:f_n(x) \to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $\sup_{t\in [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_\infty$.  So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$. 
There are other ways to see, that this can not be true. Suppose I have $f_n \to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n \to f'$. But it can definitly change pointwise (and hence also uniform) convergence. 
I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points). 
A: Since $\|f_n\|_p\to0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)
It is clear the $\lim_{n\to\infty}f_n(x)$ is $0$ if $x\in(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $x\in(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),\sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $\sup|f_n|\ge1/2>0$.
