# Determine the spectrum of the operator $A$. Is the spectrum composed only of eigenvalues of the operator $A$?

On the space $$l^2$$, define the linear operator $$A$$ with the prescription $$A(x_1,x_2,x_3,x_4,...)=((x_1+x_2)/2,(x_2+x_3)/2,(x_3+x_4)/2,...)$$.

(a) Prove that $$A$$ is a bounded operator on $$l^2$$ and calculate $$||A||$$.

(b) Determine the spectrum of the operator $$A$$. Is the spectrum composed only of eigenvalues of the operator $$A$$?

Attempt: I was able to solve part (a) and obtained $$||A||=2$$ (hopefully that's correct). However, for part (b), I'm unsure how to proceed. I know that the basic definition of the spectrum of an operator $$A$$ includes all complex numbers lambda for which the operator $$A-\lambda I$$ is not invertible. We also discussed a characterization that for $$A \in B(X)$$ holds: $$\sigma(A)$$ is equal to the union of all complex eigenvalues of the operator $$A$$ and all those complex numbers for which the operator $$A-\lambda I$$ is injective and not surjective. Therefore, I would appreciate help in solving part (b).

• The operator is of the form $A={1\over 2}I+{1\over 2} S,$ where $S$ is the shift operator. It seems $\|A\|=1.$ The spectrum of $S$ coincides with the closed unit disc. Commented May 24 at 19:33
• Oh, I see, you are right, I have made a mistake in my inequalities. But how can I from A=1/2I+1/2S justify that ||A|| is indeed 1?
– user1316777
Commented May 24 at 19:39
• First establish that both operators $I$ and $S$ separately have norm 1, and then use the triangle inequality for the operator norm. Commented May 24 at 19:49
• Ok, thanks, I agree, what about showing ||A||>=1?
– user1316777
Commented May 24 at 19:53

Now, put $$x_n=(1,1,...,1,0,...)$$, with the $$1$$s up to position $$n$$. Then $$Ax_n=(1,1,...,1,1/2,0,...)$$. So, $$\|Ax_n\|/\|x_n\|=\frac{\sqrt{n-1+1/2^2}}{\sqrt{n}}$$. This shows that $$\|A\|\geq \frac{\sqrt{n-1+1/2^2}}{\sqrt{n}}\to1$$. So, $$\|A\|=1$$.
The argument in the other answer shows that the spectrum contains the interior of the ball with center $$1/2$$ and radius $$1/2$$. To show that it is equal to the closed ball, prove that for $$r$$ outside the closure of that ball $$A-rI$$ is invertible. Then use that the spectrum is closed to conclude that it is the closed ball. To show that $$A-rI$$ is invertible, show that $$(2^{-1}-r)^{-1}A=I-\frac{2^{-1}}{r-2^{-1}}S$$ has inverse $$\sum_{k=0}^{\infty}\left(\frac{2^{-1}}{r-2^{-1}}\right)^kS^k$$. The latter is bounded since $$\|\frac{2^{-1}}{r-2^{-1}}S\|<1$$.
You can solve the system of infinitely many simultaneous equations $$\frac{x_i+x_{i+1}}2-\lambda x_i=0$$ as $$x_{i+1}=(2\lambda-1)x_i.$$ This means that the kernel of $$A-\lambda I$$ consists of (multiples of) the geometric sequence $$(2\lambda-1)^n$$ as long as that sequence is square integrable, and is trivial otherwise. Thus the spectrum is the set of all $$\lambda$$ for which $$|2\lambda-1|\leq 1,$$ i.e., a disk of radius $$1/2$$ centered at $$1/2.$$