# Linear Operators on Hilbert Spaces

1. Apr 1, 2005

### Oxymoron

Let $$U, V, W$$ be inner product spaces. Suppose that $$T:U\rightarrow V$$ and $$S:V\rightarrow W$$ are bounded linear operators. Prove that the composition $$S \circ T:U\rightarrow W$$ is bounded with $$\|S\circ T\| \leq \|S\|\|T\|$$

Last edited: Apr 1, 2005
2. Apr 1, 2005

### Oxymoron

We know that $$T$$ and $$S$$ are bounded linear operators, and that they 'operate' on three inner product spaces $$U,V,W$$. The composition is bounded if and only if it continuous. So does it make sense for me to prove that $$S \circ T$$ is continuous which implies that it is bounded?

But from what I have seen the composition of two bounded linear operators is a bounded linear operator. So what I'm trying to prove here is an obvious definition, obvious in the sense that it seems intuitive.

3. Apr 1, 2005

### Oxymoron

We know that $$S$$ is bounded. By definition it has a norm $$\|S\|$$ which is the lower bound of all the constants $$K$$ that make $$S$$ a bounded operator. ie

$$\|S\| = \inf\{K:\|Sx\| \leq K\|x\|, \, \forall \, x\in X\}$$

So I want to prove that $$S$$ is uniformly continuous if and only if $$S$$ is a bounded linear operator, say $$\|S\|= K$$.

$$proof$$

Take any two points $$v \in V$$ and $$w \in W$$, let $$d = v - w$$. Now $$\|S(d)\|$$ is bounded by $$K\|d\|$$. By linearity, $$S(d) = S(v) - S(w)$$. Thus distance is scaled by at most $$K$$, everywhere, and $$S$$ is uniformly continuous.

Conversely, take an $$r$$ such that $$\|u\| < r$$ implies $$\|S(u)\| < 1$$. The norm of the image of the ball of radius $$r$$ is at most 1, hence $$1/r$$ acts as a bound for $$S$$.

Hence $$S$$ is uniformly continuous, and by the same argument, so is $$T$$.

4. Apr 1, 2005

### Oxymoron

If I am on the right track, then since $$S$$ and $$T$$ are uniformly continuous linear operators, then $$S\circ T$$ is uniformly continuous (the composition of two continuous linear operators is again continuous and linear).

Hence $$S \circ T$$ is bounded.

5. Apr 1, 2005

### Oxymoron

Prove that if $$S,T : V \rightarrow W$$ are bounded linear operators between inner product spaces, then the map $$S + T : V \rightarrow W$$ , defined by $$(S+T)(v):= S(v) + T(v), \, \forall \, v \in V$$ , is a bounded linear operator with $$\|S+T\| \leq \|S\| + \|T\|$$.

6. Apr 2, 2005

### Oxymoron

Does anyone know if the answer to my first question is right or not?

Any pointers for my second question?

7. Apr 3, 2005

### Data

Oops, missed this! Silly physics labs.

While your strategy for the first question may work out (I haven't looked at it too hard), there is a much easier (and as you noted, intuitive!) way to go about the proof.

Say $\|T\| = K$, and $\|S\| = C$. Let $u \in U$. Then

$$T(u) = v \in V \ \mbox{with} \ \|v\| \leq K\|u\|$$

and

$$S(v) = w \in W \ \mbox{with} \ \|w\| \leq C\|v\|$$

combine the results, and see what you get

Now for the second question. Again let $\|S\| = C, \ \|T\| = K$.

Then say $S(v) = w_1 \in W, \ T(v) = w_2 \in W$, so that $\|w_1\| \leq C\|v\|$ and $\|w_2\| \leq K\|v\|$. Look at

$$(S+T)(v) = S(v)+T(v) = w_1 + w_2$$

What can be said about the magnitude of the right side (as you're probably tired of me saying, you might need the triangle inequality! )?

8. Apr 3, 2005

### Oxymoron

Thanks Data (I was wondering where you got to?). I will never tire of you reminding me to use the triangle inequality

Anyway, I'll give it a go and type back later.

9. Apr 3, 2005

### Oxymoron

Why can't $$\|v\|$$ be greater than $$K\|u\|$$ is that because we are free to choose any $$K$$ we want, and we can always choose a $$K$$ such that $$\|v\| \leq K\|u\|$$.

10. Apr 3, 2005

### Oxymoron

Let $$\|T\| = K$$ and $$\|S\| = C$$, where $$C, K \in \mathbb{R}$$.

Take any $$u \in U$$, $$v \in V$$, and $$w \in W$$. Then.

$$T(u) = v \in V$$ with $$\|v\| \leq K\|u\|$$
$$S(v) = w \in W$$ with $$\|w\| \leq C\|v\|$$

Combining the results gives

$$.\,\, \frac{1}{C}\|w\| \leq \|v\| \leq K\|u\|$$
$$\rightarrow \|w\| \leq M\|u\|$$

where $$M = CK \in \mathbb{R}$$.

That is, $$S\circ T$$ is bounded. Because, by the definition of boundedness, $$S \circ T$$ is bounded if there exists an $$M > 0 \in \mathbb{R}$$ such that $$\|(S\circ T)x\| \leq M\|x\|$$ for all $$x \in X$$.

In this case $$M = CK$$

11. Apr 3, 2005

### Data

That looks (almost - look at the next paragraph~) fine. Do you understand why $\|v\| \leq K \|u\|$ now? You can't choose $K$. It's just the order of $T$, which you know has the property that

$$\|T\| \|u\| = K\|u\| \geq \|T(u)\| = \|v\|$$

since $T(u)=v$

Your proof looks fine, except one thing. Taking any $u \in U$ was right, but you do not get to choose $v$ and $w$ arbitrarily. Since $T : U \longrightarrow V$, $v$ is just whatever vector $T$ takes $u$ to. Similarly, $w$ is whatever vector $S$ takes $v$ to. They are effectively determined by your choice of $u$. Then, you proved that $CK\|u\| \geq \|w\| = \|(S \circ T)(u)\|$, and since you could choose any $u$ to start with, you're finished!

You've even proved the last part of the question, that $\|S \circ T \| \leq \|S\|\|T\|$. Do you see why?

12. Apr 4, 2005

### Oxymoron

Thanks Data for that help and clarification. I will post back with an attempt of the second question in a couple of minutes.

13. Apr 4, 2005

### Oxymoron

By the way, regarding the first question, why does $\|S\circ T\|\leq \|S\|\|T\|$? Could you explain it please. I am really interested. In all the proofs I have seen, I actually haven't seen THIS proved to me. (don't tell me it requires the triangle inequality).

edit - please disregard this post (see below)

Last edited: Apr 4, 2005
14. Apr 4, 2005

### Oxymoron

I know why it wasn't proved to me! Because it is bloody obvious!

$$\|S\|\|T\| = CK = M$$

Hence

$$\|S\circ T\| \leq M = CK = \|S\|\|T\|$$

So $\|S\circ T\| \leq \|S\|\|T\|$

15. Apr 4, 2005

### Oxymoron

Second Question

Let $\|S\| = B$ and $\|T\| = C$, where $B, C \in \mathbb{R}$. Then take any $v \in V$ we have

$$S(v) = w_1 \in W$$ such that $\|w_1\| \leq B\|v\|$
$$T(v) = w_2 \in W$$ such that $\|w_2\| \leq C\|v\|$

Now $(S+T)(v) = S(v) + T(v)$ by linearity, and hence

$$(S+T)(v) = S(v) + T(v)$$
$$\quad = \|w_1\| + \|w_2\|$$
$$\leq B\|v\| + C\|v\|$$
$$= \|B+C\|\|v\|$$
$$= M\|v\|$$

where $M = B+C \in \mathbb{R}$. Hence by definition the map $S + T$ is bounded.

Also

$$(S+T)(v) = \|w_1\| + \|w_2\| \leq B\|v\| + C\|v\| = \|B+C\|\|v\| = \left( \|S\|+\|T\|\right)\|v\|$$

Hence

$$\|S+T\| \leq \|S\| + \|T\|$$

Last edited: Apr 4, 2005
16. Apr 5, 2005

### Data

Pretty close.

What doe you mean by

$$S(v) + T(v) = \|w_1\| + \|w_2\|,$$

though? On one side there are vectors and on the other real numbers!

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