comparison of 3 topologies on C[0,1] I have a ring of continuous functions from $[0,1]$ to $\Bbb R$. And two norms
$C[0,1]\to\Bbb R$. One is supremum of $|f(x)|,$ the other the value of
$\int_0^1|f(x)|$. Then I get a Cartesian product of continuum of sets with Euclidean topology $\prod_{t \in [0,1]}\Bbb R$ and I aim to compare these $3$ topologies. I guess the product topology is stronger than the sup one. And that sup and integral are not related. I tried to imagine the balls in sup and int topology. The first one is the function $f_0(x)$ that makes a center of a ball and every function that is between this $f_0(x) - r,$ and this $f_0(x) + r$ for every $x.$ And the integral one would be $f_0$ end every  function that makes a bulge(s) from $f_0$ with total field field of it/them no bigger than $r.$ It seems to be open in the product topology. Is it?
 A: I think it is easier to analyze the different topologies using sequences instead of open sets since all three topologies are separable.  
To begin, uniform convergence implies $L^1$ convergence but not conversely.  The first implication is a standard result in analysis, and a counter-example to the second implication is as follows.  Consider the following sequence of continuous functions: the graph of each $f_n$ is a triangle with height $1$, and the bases of the triangles are in sequence $[0,1/2]$, then $[1/2,1]$, then $[0,1/4]$, then $[1/4,1/2]$, and so on.  This sequence of functions fails to converge pointwise at any point (it takes the value $0$ and values arbitrarily close to $1$ infinitely often), so it certainly doesn't converge uniformly.  But the areas of the triangles approach $0$ as $n \to \infty$, so it does converge to $0$ in $L^1$.   
Now the product topology.  Recall that a sequence converges in the product topology if and only the projection of the sequence onto each factor converges.  The projection function $p_t \colon \Pi_t \mathbb{R} \to \mathbb{R}$ onto the $t$th factor sends a function $f$ to its value $f(t)$, so a sequence $f_n$ converges in $C[0,1]$ with the topology it inherits as a subspace of $\Pi_t \mathbb{R}$ if and only if $f_n(t)$ converges for every $t$.  In other words, this is just the topology of pointwise convergence.
It is easy to prove that a uniformly convergent sequence converges pointwise, but pointwise convergence does not imply uniform convergence as demonstrated by the sequence $f_n(x) = x^n$, for instance.  Finally, pointwise convergence and $L^1$ convergence are incomparable.  The sequence of triangular functions described above shows that $L^1$ convergence does not imply pointwise convergence.  In the other direction, let $f_n$ be the function whose graph is a triangle with height $n$ and base $[0,1/n]$.  This sequence converges to $0$ pointwise, but it does not converge to $0$ in $L^1$ because the areas of all the triangles are $\frac{1}{2}$.  Thus pointwise convergence doesn't imply $L^1$ convergence, either.
