Prove that $L^{\infty}$ and $L^{2}$ metrics on $C([a,b])$ are not equivalent.
First, I can see there are many examples which converges in $L^{2}$, not in $L^{\infty}$.
Second, for the finite interval, I can see $L^{\infty}$ convergence implies $L^{2}$.
(I think for all $p \in [1,\infty)$, $L^{\infty}$ convergence implies $L^{p}$)
But I'm not sure I can say '$L^{\infty}$ convergence implies $L^{2}$' for infinite interval.
since for the second one, $\parallel f_{n} - f\parallel_{2} \leq \parallel f_{n} - f\parallel_{\infty} \dot (b-a)^{1/2}$
Can anyone help me prove or give counter example?
Everything you've said is correct. On an infinite interval, $L^\infty$ convergence does not imply $L^2$ convergence. The constant functions $f_n = \frac 1n$ are in $C([0,\infty))$ and converge to zero in $L^\infty$, but they are not in $L^2$, so they certainly don't converge in $L^2$.
Even if you restrict to functions which are in $C([0,\infty))\cap L^2([0,\infty)) \cap L^\infty([0,\infty))$, it doesn't hold. I think I have the details right here, but certainly the idea is correct: take $$f_n(x) = \left\{ \begin{matrix} \frac 1 {n^{3/2}}\sqrt{x-n^2}, & n^2 < x \le 2n^2, \\ \frac 1 {n^{3/2}}\sqrt{3n^2-x}, & 2n^2 < x \le 3n^2, \\ 0, & \text{otherwise} \end{matrix} \right.$$ Then $\lvert f_n\rvert \le \frac{1}{n^{1/2}}$ which shows that $f_n \to 0$ in $L^\infty$. However, each $f_n^2$ has a graph that is a triangle of height $1/n$ and a base of length $2n^2$, meaning that $\|f_n\|^2_2 = n \to \infty$.