What can we say about open unit balls of sup-norm and integral-norm 
Consider the normed linear spaces $X_1=(C[0,1], ||.||_1)$ and $X_{\infty}=(C[0,1],||.||_{\infty})$ , where $C[0,1]$ denotes the vector space of all continuous real valued functions on $[0,1]$ and $$||f||_1=\int_0^1|f(t)|\,dt.$$and $$||f||_{\infty}=\sup\{|f(t)|:t\in [0,1]\}.$$Let, $U_1$ and $U_{\infty}$ be the open unit balls in $X_1$ and $X_{\infty}$ respectively. Then

*

*$U_{\infty}$ is a subset of $U_1$.


*$U_1$ is a subset of $U_{\infty}$


*$U_{\infty}=U_1$


*Neither $U_{\infty}$ is a subset of $U_1$ nor $U_1$ is a subset of $U_{\infty}$.

I really don't know from where I start to solve and conclude about open balls..Please help..

Since it seems that you want to solve it yourself, I'll just give you a hint. If you want a more accurate answer, just leave a comment.
HINT: In a normed space the unit open balls is the set of all elements which have a norm strictly less than 1. Now, what does it mean that $\left\Vert f \right\Vert_1 < 1$? What does it mean that $\left\Vert f \right\Vert_{\infty} < 1$?
Edit: We have
$$\left\Vert f\right\Vert_1 = \int_0^1 |f(t)| ~ dt \leq \int_0^1  \left\Vert f \right\Vert_{\infty} ~ dx  = \left\Vert f \right\Vert_{\infty} \cdot \int_0^1 1 ~ dx = \left\Vert f \right\Vert_{\infty}$$
for every $f \in C[0,1]$. It follows that if $\left\Vert f \right\Vert_{\infty} < 1$, then $\left\Vert f \right\Vert_1 < 1$ also. Hence, $U_{\infty} \subseteq U_1$, i.e. 1. is true.
This already shows us that 4. is false.
Now to see that 2. and 3. (which are now equivalent) are false: Consider the function
$$f(x) = \begin{cases} 2 - 8x  & , x \leq \frac{1}{4} \\ 0 &, \text{ otherwise} \end{cases}$$
Then, $f \in C[0,1], \left\Vert f \right\Vert_{\infty} = 2$ but
$$\left\Vert f \right\Vert_1 = \int_0^{1/4} 2 - 8x ~ dx = [2x - 4x^2]_{x=0}^{1/4} = \frac{1}{4} < 1$$
Hence, $f \in U_1$ but $f \notin U_{\infty}$. This shows that neither 2. nor 3. can be true.
