Finding errors in a proof that $\langle x,a\rangle\to|a|^2$ as $x$ approaches $a$

Question

I am trying to spot the mistakes in this proof where $$ \lim_{x\to a}\langle x,a \rangle = |a|^2. $$ I do not know what mistakes are made here. Thanks!

Answer 1

$\langle ., .\rangle$ is not an inner product since for $x=(1,1)$ ,$\langle x, x\rangle = 0$ whereas $x\neq0$. You may not apply the Cauchy-Schwarz inequality.

Answer 2

Note that

$$0\le|\langle x,a \rangle-|a|^2|=\left||x||a|\cos\theta-|a|^2\right|\to 0\implies \langle x,a \rangle-|a|^2\to 0$$

Answer 3

The main source of confusion is that $\langle x,a \rangle$ seems to be defined in a non-standard way as $\langle x,a \rangle = x_1a_1 - x_2a_2$, instead of the 'usual' $x_1a_1 + x_2a_2$. As noted, this has several consequences.

1) It is no longer an inner product, so Cauchy-Schwarz does not apply (see answer by Bill O'Haran).

2) It makes it unsure what $|a|^2 $ is supposed to mean at all. Normally it means $|a|^2 =\langle a,a \rangle$, which would imply $|a|^2 =a_1a_1 - a_2a_2$ (see comment by Abhinav Jha), but in the purported proof it is implied in the second line that $|a|^2 =a_1a_1 + a_2a_2$, which would be the 'normal' formula for the 'usual' definition of the dot product.

3) In the second line, the part $x_1a_1 - x_2a_2 - a_1a_1 - a_2a_2 = a_1(x_1-a_1) +a_2(x_2-a_2)$ is incorrect (the sign of the term $x_2a_2$ changes between sides).

4) Then the next part is also incorrect! Under the given definition of $\langle x,y\rangle$, we have $\langle x-a,a\rangle = (x_1-a_1)a_1 - (x_2-a_2)a_2$ (notice the different sign before $(x_2-a_2)a_2$.

That second line becomes only correct if we use the definition $|a|^2 =\langle a,a \rangle = a_1^2 -a_2^2$ and omit the faulty part in the moddle. Then it is nothing more than using that $\langle x,y\rangle$ is a bilinear form, which it still is.

Then the main error of the proof boils down to point 1).