# Norm of bounded operators of direct sum

Yesterday I posted this one regarding direct sum on Hilbert spaces $$H_1$$ and $$H_2$$ - have a look!

Direct sum of two Hilbert spaces is a inner product.

I am studying bounded operators and I just want to know if we can say something about the norm of my direct sum. I.e. let $$T_1\in B(H_1)$$ and $$T_2\in B(H_2)$$ then $$T_1\oplus T_2$$ is bounded by using the definition of bounded operator but how about its norm? I cannot se how this should be possible. Any suggestion? I belive we should use operator norm but I still cannot see how that would work. Thanks in advance.

• The operator norm of the direct sum is $\max(\| T_1 \|, \| T_2 \|)$. Try to prove it! To get some intuition you can consider first the case where $H_1, H_2$ are finite-dimensional and $T_1, T_2$ are self-adjoint (and so unitarily diagonalizable with real eigenvalues); do you know how to compute the norm in this case? Commented Nov 6, 2022 at 2:28
• So if I show that I did it in terms of $T_1$ and $T_2$? Commented Nov 6, 2022 at 2:30
• I don't know what you mean by that. Did what? Commented Nov 6, 2022 at 2:30
• Sorry. If I managed to show the operator norm of the direct sum is $\max(\| T_1 \|, \| T_2 \|)$ then I would do that in terms of the norms of $T_1$ and $T_2$? Commented Nov 6, 2022 at 2:32
• Well, there's nothing else to do it in terms of! Commented Nov 6, 2022 at 2:49

For any $$(x,y)\in H_1\oplus H_2$$, by definition, $$T_1\oplus T_2 (x,y)=(T_1 x,T_2 y$$). Thus \begin{align*}\|T_1\oplus T_2(x,y)\|=\sqrt{\|T_1x\|_1^2+\|T_2y\|_2^2}&\leq \sqrt{\|T_1\|_1^2\|x\|_1^2+\|T_2\|_2^2\|y\|_2^2}\\&\leq\max\{\|T_1\|_1,\|T_2\|_2\}\|(x,y)\|.\end{align*} This gives an upper bound for what the norm of $$T_1\oplus T_2$$ is (and shows that it is definitely bounded at least). To show that this bound is the norm, I leave you the fun of choosing an appropriate $$(x,y)$$.

• Hi, thank you for your comment! Can you explain why you use subsript ${\color{Red} 1}$ for $\|T_1x\|_{\color{Red} 1}^2$ and subscript ${\color{Red} 2}$ for $\|T_2y\|_{\color{Red} 2}^2$? I see that this is bounded which I already have shown, so to find such an $(x,y)$. Can you give me an additional hint? Commented Nov 6, 2022 at 14:43
• @TheCalcLover1231 The subscript just indicate that the norm is taken in the appropriate space (as these norms could be different). To show that an upper bound is the least upper bound (and thus the norm) it suffices to shows that this bound can be achieved by some vector (x,y). You thus need to find an $(x,y)$ such that the inequality above is actually an equality. Commented Nov 6, 2022 at 19:19