discontinuity of an operator I want to show that if $X=Y$ is the subspace of $L^1(0,1)$ over $\mathbb C $ consisting of all polynomials, then $T:X \times Y \rightarrow \mathbb C$ given by
$T(f,g)=\int_0^1 f(t)g(t) dt$
is not continuous.
I have tried the following:
$T$ is surely bilinear, hence if it is continuous it is bounded.
I am not given the norm in $X\times Y$ but I then proceed saying take $x_n=(n+1)x^n$ 
$|T(x_n,x_n)|\rightarrow \infty $ for $n\rightarrow \infty$
But if T were bounded, $|T(x_n,x_n)|\leq M ||(x_n,x_n)||_{X\times Y}$
now I would imagine we can bound the norm in the product space by a sum of the norms in $L^1$ but I am not sure how to go about it.
If we can get this bound then we are done since $||x_n||_X=1 \ \ \ \forall n$ 
If you could help filling the gap or find a mistake in my reasoning it would be extremely helpful!
 A: For normed spaces $X,Y$, the continuity or boundedness of a bilinear form
$$T \colon X\times Y \to \mathbb{K}$$
is equivalent to the existence of a constant $M \geqslant 0$ with
$$\lvert T(x,y)\rvert \leqslant M\cdot \lVert x\rVert\cdot \lVert y\rVert\tag{1}$$
for all $x\in X$, $y\in Y$. That $(1)$ implies continuity is easily verified, and conversely, by continuity, there is  a neighbourhood $W$ of $(0,0)$ such that $\lvert T(x,y)\rvert  \leqslant 1$ for all $(x,y) \in W$. By the definition of the product topology, there are $\alpha, \beta > 0$ such that $B_\alpha \times B_\beta \subset W$, where $B_r$ is the (closed, but that's not important) ball with radius $r$ and centre $0$ in the respective space. Then one can verify that
$$\lvert T(x,y)\rvert \leqslant \frac{1}{\alpha \beta}\lVert x\rVert\cdot \lvert y\rVert$$
for all $x,y$.
Thus, the chosen $x_n(t) = (n+1)t^n$ show that the given $T$ is not continuous.
A: You are right, the norm on the product space is not given, so one would assume that it is a $p$-norm:
$$
|| (x,y) || =\left(||x||^p + ||y||^p\right)^{\frac 1 p}
$$
which are all equivalent each other. Hence you can assume $p=1$ even if usually $p=2$ is the preferred choice.
