so the definition for a bounded linear operator $F:X \rightarrow Y$ is that $$||F(x)||_X \leq C||x||_Y$$ for some constant $C$ and all $x$. So my question: is the $C$ dependent on the choice of $x$?
For example I am thinking if $F(x_1,x_2) = (x_1,x_1+x_2)$ that is linear. Then $||F(x_1,x_2)|| = || (x_1,x_1+x_2)|| \leq ||(x_1,x_2)|| + ||(x_1,x_2)||$. Then because $C$ is independant of $x$ we cannot have $ ||(x_1,x_2)|| + ||(x_1,x_2)|| \leq C||(x_1,x_2)|| $, correct? Then this $F$ is not bounded.
If you are using the Euclidean norm in $\mathbb R^2$, then \begin{align} \|(x_1,x_1+x_2)\|&\leq \|(x_1,x_1)\|+\|(x_1,x_2)\|=|x_1|\,\|(1,1)\|+\|(x_1,x_2)\|\\ \ \\ &=\sqrt2\,|x_1|+\|(x_1,x_2)\|\\ \ \\ &\leq\sqrt2\,\|(x_1,x_2)\|+\|(x_1,x_2)\|\\ \ \\ &=(\sqrt2+1)\,\|(x_1,x_2)\|. \end{align} So you can take $C=\sqrt2+1$. The actual norm of $F$ is a bit smaller; it is not hard to check that $$\|F\|=\frac{1+\sqrt5}2.$$