Give an example of a discontinuous bilinear form. I'm having a hard time with this, I asked a tutor and he couldn't think of an example either.
The bilinear form on a vector space $V$ is a map $f:  V \times V \rightarrow F$ that satisfies linearity in both variables.
 A: In order to have such an example, you need $V$ to be infinite-dimensional. Here’s a concrete example.
Let $F=\mathbb{R}$, and let $V$ be the set of infinite sequences of real numbers that are eventually $0$ (i.e., that have only finitely many non-zero terms) with the uniform (or sup) norm. $V$ is a subspace of $\ell^\infty$, and in fact of $c_0$. Now define $$f:V\times V\to\mathbb{R}:\langle x,y\rangle\mapsto\sum_{n\in\mathbb{N}}2^nx(n)y(n)\;;$$ for each $\langle x,y\rangle\in V\times V$ only finitely many terms of the sum are non-zero, so $f$ is well-defined, and it’s obviously bilinear.
For $n\in\mathbb{N}$ define $x_n\in V$ by $$x_n(k)=\begin{cases}2^{-n},&\text{if }k=n\\0,&\text{if }k\ne n\end{cases}\;,$$ and let $z$ be the zero vector. Clearly $$\Big\langle\langle x_n,z\rangle:n\in\mathbb{N}\Big\rangle\to\langle z,z\rangle$$ in $V\times V$, since $\|x_n\|_\infty=2^{-n}$, but $f\big(\langle x_n,z\rangle\big)=1$ for every $n\in\mathbb{N}$.
A: Alternatively, if you let $P$ be the space of real-valued polynomials of a real variable s.t. $\int_0^1 | p | < \infty$ for all $p \in P$, then the map $P \times P \to \mathbb{R}$ given by $(p,q) \mapsto \int_0^1 pq$ is bilinear, but discontinuous. In particular, if we choose $p_n = q_n = \sqrt n x^n$, then $p_n \to 0$ but $(p_n, p_n) \mapsto 1/2$ in limit.
You could be more lenient, and choose $V$ just to be the space of real valued continuous functions whose L1 integral is finite, and consider the same map $V \times V \to \mathbb{R}$. It's easy to see that this one isn't continuous, as now you can choose bump-style functions that converge non-uniformly to 0 everywhere, but whose product doesn't.
Again, these are both infinite dimensional.
A: I am assuming that $V$ is an infinite-dimensional normed space. Let $(e_i)_{i\in I}$ be a Hamel-basis of $V$, we may assume that  $\| e_i\|=1 $ for all $ i\in I$. Now, $(e_i\otimes e_j)_{i,j\in I} $ is a basis for the tensor product $V\otimes V$. Pick an infinite sequence $(i_n,j_n)$ of elements of $I^2$. Now, we can define  $f(e_{i_n}\otimes e_{j_n})=n$  and define it to be zero on the rest of the basis and then extend linearly to a map on $V\otimes V$, by the universal property of the tensor product this induces a bilinear $F:V\times V\to \mathbb R$. 
It should be clear that $F$ is discontinuous.     
