The bilinear functional cannot be continuous on the space X In the below exercise , the following argument I write is wrong, but I wondered to see what point I miss here. Also, I could not think of a counter example.
 
since both functions $p,q$ are polynomials they are continuous and hence on the set $[0,1]$ they attain their maximum. let $\max_t |p(t)|= M <\infty$ and $\max_t |q(t)|= N <\infty$ so ;
$$|B(p,q)| \le \int_0^1 |p(t)q(t)|dt \le M.N <\infty$$
 A: Your argument shows nothing: you already know that $B(p,q)$ is a number, so showing that its absolute value is less than another number does not contribute anything. Here "bounded" (=continuous) as a bilinear form would mean that there exists $c>0$, independent of $p,q$,  such that 
$$\tag1
|B(p,q)|\leq c\|p\|\,\|q\|.
$$
So basically we need to see that the integral of the product can be big even if the integral of each factor isn't. To look for a counterxample, let us try something simple: let $p(t)=q(t)=t^n$. Then 
$$
|B(p,q)|=\int_0^1t^{2n}\,dt=\frac1{2n+1},
$$
while 
$$
\|p\|=\int_0^1 t^n\,dt=\frac1{n+1}.
$$
So if $B$ is bounded, we would have $(1)$ for some $c>0$:
$$
\frac1{2n+1}\leq c\left(\frac1{n+1} \right)^2=\frac{c}{(n+1)^2},
$$
which is impossible for all $n$. So $B$ is not bounded. This suggests how to write the example if you prefer to talk about continuity instead of boundedness: let $p_n(t)=\sqrt{n}\,t^n$. Then 
$$
\|p_n\|=\int_0^1 \sqrt{n}\,t^n\,dt=\frac{\sqrt{n}}{n+1},
$$
so $p_n\to0$. Meanwhile, 
$$
B(p_n,p_n)=\int_0^1 n t^{2n}\,dt=\frac{n}{2n+1}\to 1.
$$
So $$\lim_{n\to\infty}B(p_n,p_n)\ne B(\lim_{n\to\infty}(p_n,p_n)),$$
showing that $B$ is not continuous. 
