I am not sure about the last step of my proof:
$(L^{p}(X,A,\mu), \|\cdot \|)$ is a normed $L^{p}$ space of p-integrable functions. $L^{p}(a,b)$ is the space of p-integrable functions on (a,b). $C[a,b]$ is the space of continuous functions on [a,b].
Main goal in order to prove the claim is that the integral of $|f|^{p}$ is finite for any $f\in C[a,b]$. Let us take a function $f\in C[a,b]$. Now we see that the interval $[a,b]$ is compact (assuming both $a$ and $b$ are finite real numbers). Therefore, $f$ has a maximum. Similarly, the function $|f|^{p}$ is also continuous on $[a,b]$ and hence has maximum on it. Now the integral of the function $|f|^{p}$ is dominated by $\max|f|$ (max on the interval $[a,b]$) times $(b-a)$ (the Lebesgue size of the interval). And therefore it (the integral which I just mention) is finite. And this inequality shows that $C[a,b]$ is a subspace of $L_{p}$ for any $1\le p\le \infty$. And it is linear because $\forall f,g \in C[a,b]$ and $a,u$ $\in \mathbb{R}$ $af+ug \in C[a,b]$. It remains to be checked whether the linear subspace is indeed closed.
My friend told me about this idea, but that still didn't help me: "Since all the step functions are dense in $L_p(a,b)$ and we know that any step function can be approximated by continuous functions on [a,b]."
Any help would be greatly appreciated!
Thanks
Well, you could have done the shortcut: if $f$ is continuous over $[a,b]$, then so is $|f|^p$, so it is integrable over the compact interval $[a,b]$.
However, it is not closed. Consider for simplicity $[a,b]=[-1,1]$ and let $f$ be the signum function on it: $f(x)=\displaystyle\frac x{|x|}$ if $x\ne 0$ and $f(0)=0$. This is not continuous, but can be approximated arbitrarily close (even in $p$-norm) by continuous functions.
(Probably the simplest is to define $f_n$ as the piecewise linear function satisfying $f_n(x)=f(x)$ if $|x|>1/n$ and connects the points $(-1/n,\,-1)$ and $(1/n,\,1)$ by a straight line.)