I wanted to know if my proof is really a contradiction or not. My TA says it is, but I am still skeptical based on the definition of a contraction that we use. Here is the statement of the problem:
Consider the linear function $f:\mathbb{R^{2}}\rightarrow\mathbb{R^{2}}$ defined by $$f\left(x\right)=\frac{K}{\sqrt{2}}\left(x_{1}+x_{2},x_{2}-x_{1}\right), \forall x=\left(x_{1},x_{2}\right)\in\mathbb{R^{2}}$$ Show that when the $1$-norm is used (i.e., $\|\cdot\|_{1}$), $f$ is not a contraction if $K\in\left(\frac{1}{\sqrt{2}},1\right)$.
Previously, I had shown that if $f$ is a linear self-mapping with the metric $d\left(x,y\right)=\|x-y\|$ for $x,y\in V$, where $\left(V,\|\cdot\|\right)$ is a complete normed space, then $f$ is a contraction if and only if there exists a constant $C$ with $C\in\left(0,1\right)$ such that $\|f\left(x\right)\|\le C\|x\|$.
I tried to use this in a proof by contradiction by supposing that $f$ is a contraction, then $$\|f\left(x\right)\|_{1}\le C\|x\|_{1}\implies \frac{K^{2}}{2}\left(\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}-x_{1}\right)^{2}\right)\le C\left(x_{1}^{2}+x_{2}^{2}\right)\implies K^{2}\le C$$
I am not sure if I have reached a contradiction yet. I don't believe I have, but my TA said that the contradiction is that $C$ is unbounded, but I don't see how this is a contradiction because by our definition of a contractive function, we just have to find one $C\in\left(0,1\right)$ that satisfies this inequality, so any $C\in\left[K^{2},1\right)$ will work. Is the actual contradiction that $C$ is unbounded above?
No, you have not reached a contradiction. The big problem I see is that you are not using the $1$-norm at all; $\frac{K^2}{2} \Big( (x_1 + x_2)^2 + (x_2 - x_1)^2 \Big)$ is the square of the $2$-norm.
Also, to prove by contradiction, it is better to pick a concrete value of $x$ (a counterexample). For example, let $x = (0,1)$; then $$\|f(x)\|_1 = \frac{2}{\sqrt{2}}K,$$ which should lead to a proof.