Bilinear forms on C[0,1] Let $C[0,1]$ be the vector space of real-valued continuous functions on $[0,1]$. Then
$$B(f,g) = \int_0^1{f(x)g(x)\, dx}$$
is a bilinear form on $C[0,1]$. More generally, if $k:[0,1]^2\rightarrow \mathbb{R}$ is continuous, then
$$B_k(f,g) = \int_{[0,1]^2}{f(x)g(y)k(x,y)\,dx\, dy}$$
is a bilinear form.
In Keith Conrad's notes on bilinear forms, he asks if there is a $k$ corresponding to the first example. My intuition first suggested the characteristic function on the diagonal, but that is not continuous. Surely it must be something like this, but I can't think of what it should be.
There is another question where he asks us to find conditions on $k$ under which $B_k$ will be a symmetric bilinear form. I sketched a proof that its both necessary and sufficient for $k(x,y) = k(y,x)$, but my proof is not pretty. Is there a natural proof?
 A: For the first question: we would have for any $f$ and $g$ continuous that $\int_0^1f(x)\left(g(x)-\int_0^1k(x,y)g(y)dy\right)dx$. This implies that $g(x)=\int_0^1k(x,y)g(y)dy$ for all $g$ continuous. By Stone-Weierstrass theorem, $k$ is approximable by finite sums of terms of the form $a(x)b(y)$, hence the identity from $[0,1]$ to $[0,1]$ would be compact: it's not possible. 
In $B_k(g,f)$, switch the role of $x$ and $y$: this gives for each $f$ and $g$ continuous,
$$\int_{[0,1]^2}f(x)g(y)(k(x,y)-k(y,x))dxdy=0.$$
Take $F(x,y)$ continuous on $[0,1]$ and approximate by linear combinations $\sum a_i(x)b_i(y)$.
A: 1) No. If such a $k$ existed, denote $M:=\sup |k|$. Then
$$
\big|\int_0^1 f^2\Big|=\Big|\iint f(x)f(y)k(x,y)dxdy\big|\leq \iint|f(x)||f(y)|Mdxdy=M\left(\int_0^1 |f|\right)^2
$$
for every continuous function $f$ on $[0,1]$. Then for each $n\geq 1$, consider the continuous function $f_n$ defined by $f_n(x):=\frac{1}{\sqrt{x}}$ on $[1/n,1]$ and $f_n(x):=\sqrt{n}$ on $[0,1/n]$. We get
$$
\log n=\int_{1/n}^1\frac{1}{x}dx\leq \int_0^1f_n^2\leq M\left(\int_0^1 |f_n|\right)^2\leq M\left(\int_0^1 \frac{1}{\sqrt{x}}dx\right)^2=4M.
$$
Let $n$ tend to $+\infty$ to obtain the desired contradiction.
2) If $B_k(f,g)=B_k(g,f)$, set $u(x,y):=k(x,y)-k(y,x)$. You get, as observed by Davide Giraudo, 
$$
\iint_{[0,1]^2} f(x)g(y)u(x,y)dxdy=0
$$
for every $f,g$ continuous on $[0,1]$. From here, there are several ways to conclude that $u=0$. Here are two possible ones.
a) Applying the latter to $f(x)=x^n$ and $g(y)=y^m$, we get, by linearity, $\iint p(x,y)u(x,y)dxdy=0$ for every polynomial in two variables $p(x,y)$. By Stone-Weierstrass, $u$ can be uniformly approximated by such polynomials. So $\iint u^2=0$ whence $u^2=0$ since $u^2$ is continuous and nonnegative.
b) Approximating uniformly characteristic functions of intervals by continuous functions, we get $\iint 1_I(x)1_J(y)u(x,y)dxdy=\iint 1_{I\times J}\cdot u=0$ for every intervals $I,J$ in $[0,1]$. If $u$ was not $0$, by continuity, there would exist non trivial intervals $I,J$ such that $u\geq \alpha>0$ on $I\times J$ (or $u\leq -\alpha<0$). Then $0=\iint 1_{I\times J}\cdot u\geq \alpha \lambda(I)\lambda(J)>0$ (or $<0$). Contradiction.  
