Prove $\|f\|_{L^p}$ is not equivalent to $\|f\|_{\infty}$ in $C[a,b]$

Question

Prove that in $C[a,b]$ the uniform norm is not equivalent to the $L^p$ norm for $(1\leq p < \infty)$ I am stuck on showing that the function below satifies the claim. I know that f is continuous and the uniform norm gives a finite result. Im not sure how to proceed. In the normed linear vector space $(C[a,b],\|.\|)$ we have to find a function where the norms are not equivalent, i.e.: $\nexists C,c C>c>0$ such that $c\|f\|_{L^p}\leq \|f\|_{\infty} \leq C\|f\|_{L^p}$ $$\|f\|_{L^p}= \left(\int_a^b \lvert\,f(t)\rvert^p d \mu\right)^{\frac{1}{p}}, \quad \|f\|_\infty = \sup_{x \in [a,b]}\lvert f(x)\rvert.$$ Take $f_n(x) =\left\{ \begin{array}{ll} 1-nx & \mbox{on } x \leq 1/n \\ 0 & \mbox{on } x > 1/n \end{array} \right.$

thanks for any help!

Answer 1

Another example, with functions $f_n$ in $C^\infty([0,+\infty))$: $$ f_n(x)=\mathrm e^{-nx}.$$ Then $\|f_n\|_\infty=1$ and $\|f_n\|_p=(np)^{-1/p}\to0$ hence the inequality $\|f\|_\infty\leqslant C\|f\|_p$ for every $f$ in $C^\infty([0,+\infty))$ and for some finite constant $C$ independent of $f$, is impossible.

Answer 2

Take $f(x)=|x-a|^{-1/2p}$. Clearly $f\in L^p[a,b]$, but $f\not\in L^\infty[a,b]$. Then take $$ f_\delta(x)=\left\{ \begin{array}{lll} f(a+\delta) & \text{if} & x\in [a,a+\delta), \\ f(x) & \text{if} & x\in [a+\delta,b]. \end{array} \right. $$ Clearly, $f\delta$ belongs to both $L^p[a,b]$ and $L^\infty[a,b]$, and $$ \|f_\delta\|_p\le\|f\|_p=\left(2(b-a)^{1/2}\right)^{\!1/p}, $$ while $$ \|f_\delta\|_\infty=f_\delta(a)=\delta^{-1/2p}, $$ and hence there is no $c$ such that $$ \|f_\delta\|_{\infty}\le c\|f_\delta\|_{p}. $$