Comparison of norms over a vector space of continuous functions Trying to compare one-norm, two-norm and infinity norm over C[0,1]. We are given the following:
For f $\in C[0,1]:$
$||f||_1 = \int\limits_0^1 |f(t)|dt, \hspace{2.5cm} ||f||_\infty = max_{t\in[0,1]} | f(t)|, \hspace{2.5cm} ||f||_2 = \sqrt{\int\limits_0^1 f(t)^2dt}$  
The question is how can we prove that  $ ||f||_1 \leq ||f||_2 \leq ||f||_\infty $ for all $f \in C[0,1]$?
It makes sense to me that, since the one norm gives the area between the f(t) curve and the t axis, and the infinity norm gives the maximum distance between f and the t-axis, that $ ||f||_1 \leq ||f||_\infty $ over [0,1]. But the relationship of the two with $ ||f||_2 $ is what we really need to compare in this case, and it happens to be what I have failed to wrap my head around.
 A: In order to prove $||.||_1\leq ||.||_2$, use Hölder's inequality. See also the answer in this discussion (same method as in the finite dimension case). 
There is a similar question there (i.e. you can use the same technics). If you find it unclear, do not hesitate to interact. 
You can also use the Cauchy Schwartz inequality which I reproduce here. Let $f,g\in C([0,1])$ and define the (Hermitian) scalar product
$$
\langle f|g\rangle:=\int_0^1 f(t)\bar{g}(t) dt
$$
then $\langle f|f\rangle=(||f||_2)^2$. Now, consider, still for $f,g\in C([0,1])$ and $s\in \mathbb{R}$, the quadratic function 
$$
(||f+s.g||_2)^2=(||f||_2)^2+2\,s.\Re(\langle f|g\rangle)+s^2.(||g||_2)^2
$$
from the fact that this quadratic function is always positive, one gets that its discriminant 
$$
\Delta=4.(||f||_2||g||_2)^2-\Re(\langle f|g\rangle)^2)
$$
is negative. Hence
$$
|\Re(\langle f|g\rangle)|\leq (||f||_2||g||_2)
$$
which still holds true even if $||g||_2=0$. 
Now, let $u\in C([0,1])$. With $|u|=f$ and $g=1_X$ (the unity constant function on $X=[0,1]$, the support of the measure space which is of measure $1$) one gets 
$$
||u||_1=|\Re(\langle f|g\rangle)|\leq (||f||_2||g||_2)\leq ||u||_2
$$ 
which is the desired inequality. To end the proof one remarks that, for $t\in [0,1]$, $|u(t)|\leq ||u||_\infty. 1_X$, integrates and obtains 
$$
||u||_2\leq ||u||_\infty
$$  
