Does $\lim_{y\to x}\frac{|B(y_1-x_1,y_2-x_2)|}{\|(y_1-x_1,y_2-x_2)\|}$ always approach 0 for a bilinear map $B$?

Question

im trying to show that the total derivative of a bilinear map $B:X^2\to\mathbb{R}$ (where $X$ ist some vectorspace) is $dB(x)(h)=B(x_1,h_2)+B(h_1,x_2)$ by using the limit definition $$\lim_{y\to x}\frac{|B(y)-B(x)-dB(x)(y-x)|}{\|(y_1-x_1,y_2-x_2)\|}=0$$ where $\|\cdot\|$ ist some norm of $X^2$. Using the properties of bilinear functions I end up with $\lim_{y\to x}\frac{|B(y_1-x_1,y_2-x_2)|}{\|(y_1-x_1,y_2-x_2)\|}$. Somehow I can't figure out why this should approach $0$. Can someone help me?

Answer 1

The conclusion is not true unless $X$ is finite dimensional.

Let $\varphi(x)$ be a discontinuous linear functional on the infinite dimensional Hilbert space $\mathcal{H}.$ Then $$B(x,y)=\varphi(x)\varphi(y)$$ is a bilinear map. There exists a sequence $x_n$ such that $x_n\to 0$ and $|\varphi(x_n)|\ge 1.$ Then $$|B(x_n,x_n)|=|\varphi(x_n)|^2\ge 1,\qquad (x_n,x_n)\to (0,0)$$