# Proving a linear operator is always bounded

Give an example of a linear operator $T: \mathbb{R^2} \rightarrow \mathbb{R^2}$ (choose whatever norms in $\mathbb{R^2}$ that you like). Show that your linear operator is bounded always.

We never went too much into linear operators, so I understand the concept here but not exactly how the proof would operate.

For example, such an operator could look like $T(x_1, x_2) = (y_1, y_2)$, correct? Or, for any set of two points in $\mathbb{R^2}$, $T$ gives us two different points.

However, I'm not sure what the form of such an operator would look like. I can imagine $T: \mathbb{R^2} \rightarrow \mathbb{R}$ (a simple function of 2 variables), but I'm not sure how this works with an output in $\mathbb{R^2}$.

Also, I'm not sure how to prove boundedness - is it true that any such $T$ is always bounded, or do you have to choose a $T$ which is bounded?

• well, you have to say what $y_1$ and $y_2$ are, or how they are depending on $x_1,x_2$. Hint: It does need to be very complicated... – user251257 Apr 18 '17 at 16:59
• @user251257 My question has a lot to do with the definition of a linear operator ($A(x_1 + x_2) = Ax_1 + Ax_2$, and $A(cx) = cAx$.) I'm not sure how this works for a linear operator plotting $\mathbb{R^2}$ into $\mathbb{R^2}$. For example, if we had some operator $T$ such that $T(x, y) = (2xy, x + y)$, how do we use the definition to show that this is a linear operator? (I'm not sure that it is) – mizichael Apr 18 '17 at 17:07
• I wanted to write "it doesn't need to be very complicated". Sorry for the confusion. The definition for linear ist just $T(ax + by) = aT(x) + bT(y)$ for every $a,b\in\mathbb R$ and $x,y\in\mathbb R^2$. Now, for your $T$: Can $(x,y)\mapsto xy$ be linear? – user251257 Apr 18 '17 at 17:09
• No, I suppose it can't be linear. Going back to your definition for a linear operator, however, I'm confused about the operator $T$. When you plug in some set of $x, y\in \mathbb{R^2}$ to $T$, it gives you some set of numbers $c, d\in \mathbb{R^2}$, right? It seems that $T(ax + by)$ is really $T: \mathbb{R} \rightarrow \mathbb{R^2}$ because $ax + by$, for any values you choose, is really just a number in $\mathbb{R}$. – mizichael Apr 18 '17 at 17:18
• well, you used $x,y$ as numbers not as vectors. I see it is kind of confusing. I will just post an answer. – user251257 Apr 18 '17 at 17:20

## 1 Answer

There are two pretty simple linear operators on any vector space, namely the zero operator: $$\mathbb R^2 \ni x \mapsto T_0(x) := 0$$ and the identity operator: $$\mathbb R^2 \ni x \mapsto T_1(x) := x.$$ Any norm of $T_0$ must be $0$, by definition. So that is pretty boring. In the usual operator norm / supremum norm we have $\|T_1\| = 1$, as $$\sup_{x\ne 0} \frac{\|T_1(x)\|}{\|x\|} = \sup_{x\ne 0} \frac{\|x\|}{\|x\|} = 1.$$

Now, for a not so boring example, for fixed $a,b,c,d\in\mathbb R$ consider: $$\mathbb R^2 \ni (x_1, x_2) \mapsto T(x) := \begin{bmatrix} ax_1 + bx_2 \\ cx_1 + dx_2 \end{bmatrix}.$$ Let $\|x\|$ denote the usual euclidean norm. Then, the Cauchy Schwarz inequality yields $$\|T(x)\|^2 = (ax_1 + bx_2)^2 + (cx_1 + dx_2)^2 \le (a^2 + b^2)\|x\|^2 + (c^2 + d^2)\|x\|^2 \le (a^2 + b^2 + c^2 + d^2)\|x\|^2.$$ That is, $$\| T \| = \sup_{x\ne 0} \frac{\|T(x)\|}{\|x\|} \le \sqrt{(a^2 + b^2 + c^2 + d^2)}.$$ Notice, that upper bound is not optimal. To obtain $\|T\|$, one usually needs some linear algebra.

• This makes a lot more sense, thank you very much! Is it true that $||T|| \leq ||x|| \sqrt{(a^2 + b^2 + c^2 + d^2)}$? Thus, if we simply choose $||x||$ arbitrarily large, how could $T$ possibly be bounded? – mizichael Apr 18 '17 at 17:39
• @mizichael: No, we have $\|T(x)\| \le \|x\| \sqrt{a^2 + b^2 + c^2 + d^2}$ and $\|T\| \le \sqrt{a^2 + b^2 + c^2 + d^2}$. Please don't mix them. Also a bounded linear operator means, that $\|T\| < \infty$. $\|T(\mathbb R^2)\|$ is always unbounded if $T\ne 0$. – user251257 Apr 18 '17 at 17:43
• My mistake, I was a bit confused by the notation. Just to clarify, what exactly is the distinction between $T$ and $T(x)$? – mizichael Apr 18 '17 at 17:44
• @mizichael $T$ is the operator, a function. $T(x)$ is the value of $T$ at $x$, a vector. – user251257 Apr 18 '17 at 17:45
• So what exactly does it mean for the function $T$ to be bounded while the value of $T$ at $x$, a vector, is unbounded? I'm not sure I understand how that can be. – mizichael Apr 18 '17 at 17:52