Why in the defn of bounded linear functional does the bound depend on $x$? If $T : X \to Y$ is a linear functional between normed spaces, we say $T$ is bounded if $\exists M > 0$ such that $||T(x)||_{Y} \leq M ||x||_{X}$ for all $x \in X$.
Usually, when we say bounded, e.g., for $f : \Bbb R \to \Bbb R$, we mean there is some $M > 0$ such that for all $x \in \Bbb R$, $|f(x)| \leq M$.  Here, the right hand side (RHS) doesn't depend on $x$.  
But for some reason in the first definition, our RHS depends on $x$.  What is the reason for this?
Edit: If I remember correctly, we say a map $f : D_{1} \to D_{2}$ between metric spaces is bounded if there is some $M > 0$ such that $d_{2}(f(x),f(y)) \leq M d_{1}(x,y)$ for all $x, y \in D_{1}$.  If this is the correct definition, then my question applies to this scenario, too.  Why does the RHS depend on the input?
 A: The reason is that for linear functions on normed spaces, the only functional that is bounded in the usual sense is the zero functional.
Linear is so much more restrictive than say continuous, or even smooth or analytic, that if we also impose the usual definition of boundedness, there is nothing interesting left to study.
A: If you like, you could rephrase this condition as saying that the map is bounded on the unit sphere.
A: If we required $\forall x\in X\quad\|Tx\|_Y\leq M$, there would be no linear map satisfying this condition. 
A: A linear function on a normed linear space is continuous everywhere iff it is continuous at the origin. And it is continuous at the origin if, for every $\epsilon > 0$, there exists $\delta > 0$ such that
$$
                    \|x\| < \delta \implies \|Tx\| < \epsilon.
$$
For all $0 < r < 1$, and $x \ne 0$,
$$
              \left\|\frac{r\delta}{\|x\|}x\right\|< \delta \implies
         \left\|T\left(\frac{r\delta}{\|x\|}x\right)\right\| < \epsilon
$$
Therefore,
$$
                   \|Tx\| < \frac{\epsilon}{r\delta}\|x\|,\;\;\; x \ne 0, 0 < r < 1.
$$
Letting $r\rightarrow 1$ gives
$$
                       \|Tx\| \le \frac{\epsilon}{\delta}\|x\|.
$$
Conversely, if $\|Tx\| \le M\|x\|$ for all $x$, then
$$
                \|x\| < \frac{\epsilon}{M+1} \implies \|Tx\| \le M\|x\| < \frac{\epsilon M}{M+1} < \epsilon.
$$
So boundedness of a linear map is equivalent to continuity at $0$, which is equivalent to continuity at every point.
