## Linear Operators on Hilbert Spaces

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 $$\|S\circ T\| \leq \|S\|\|T\|$$
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 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.
 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 $$T$$.

## Linear Operators on Hilbert Spaces

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.
 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 $$\|S+T\| \leq \|S\| + \|T\|$$.
 Does anyone know if the answer to my first question is right or not? Any pointers for my second question?
 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! )?
 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.
 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 $$\|v\| \leq K\|u\|$$.
 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 $$M = CK$$
 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?
 Thanks Data for that help and clarification. I will post back with an attempt of the second question in a couple of minutes.
 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)
 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 $\|S\circ T\| \leq \|S\|\|T\|$
 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 $$\|S+T\| \leq \|S\| + \|T\|$$
 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!