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Let $H$ be a Hilbert space and let ${e_n} ,\ n=1,2,3,\ldots$ be an orthonormal basis of $H$. Suppose $T$ is a bounded linear oprator on $H$. Then which of the following can not be true? $$(a)\quad T(e_n)=e_1, n=1,2,3,\ldots$$ $$(b)\quad T(e_n)=e_{n+1}, n=1,2,3,\ldots$$ $$(c)\quad T(e_n)=e_{n-1} , n=2,3,4,\ldots , \,\,T(e_1)=0$$

I think $(a)$ is not true because $e_1$ can not span the range space. I really don't know how to approach to this problem. Could you please give me some hints? Thank you very much.

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  • $\begingroup$ Why couldn't $e_1$ span the range space? $\endgroup$ – Eric Wofsey Sep 14 '18 at 17:31
  • $\begingroup$ Because it is singleton set. I may be wrong. $\endgroup$ – Ziya Sep 14 '18 at 17:32
  • $\begingroup$ What does being a singleton have to do with anything? $\endgroup$ – Eric Wofsey Sep 14 '18 at 17:33
  • $\begingroup$ Could you please give me some hints? $\endgroup$ – Ziya Sep 14 '18 at 17:34
  • $\begingroup$ Is the linear functional with $e_k\mapsto1$ bounded? $\endgroup$ – Lord Shark the Unknown Sep 14 '18 at 17:36
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$(a)$ cannot be true. Assume that such $T$ exists. Then recall that $\sum_{n=1}^\infty \frac1n e_n \in H$ so

$$T\left(\sum_{n=1}^\infty \frac1n e_n\right) = \sum_{n=1}^\infty \frac1n Te_n = \left(\sum_{n=1}^\infty \frac1n\right)e_1$$

but the latter sum doesn't converge. This is a contradiction.

$(b)$ and $(c)$ can be true. Consider $$Sx = \sum_{n=2}^\infty \langle x, e_{n-1}\rangle e_{n}$$

and

$$Tx = \sum_{n=1}^\infty \langle x, e_{n+1}\rangle e_{n}$$ Both are contractive and hence bounded.

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  • $\begingroup$ What if$ T(e_n)=\sqrt(\frac{n+1}{n})e_n$ or some other number into $e_n?$ $\endgroup$ – Ziya Sep 15 '18 at 3:46
  • $\begingroup$ @Ziya The only candidate for $T$ is given by $$Tx = T\left(\sum_{n=1}^\infty\ \langle x, e_n\rangle e_n \right) = \sum_{n=1}^\infty \langle x, e_n\rangle Te_n = \sum_{n=1}^\infty \sqrt{\frac{n+1}n}\langle x, e_n\rangle e_n$$ We have $$\|Tx\|^2 = \sum_{n=1}^\infty \frac{n+1}n |\langle x, e_n\rangle|^2 \le 2 \sum_{n=1}^\infty |\langle x, e_n\rangle|^2 =2\|x\|^2$$ so $T$ is well-defined and bounded. $\endgroup$ – mechanodroid Sep 15 '18 at 7:16
  • $\begingroup$ Thank you very much. $\endgroup$ – Ziya Sep 15 '18 at 9:26
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Let's look at (a).

If $T$ satisfies $T(e_n)=e_1$ for all $n$, then $$T(e_1+e_2+\cdots+e_n)=ne_1.$$ But $\|e_1+e_2+\cdots+e_n\|=\sqrt n$ and $\|ne_1\|=n$.

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You're questions been answered, but if you're curious about examples of bounded linear operators that satisfy the last two resp. Look no further than $l^2(\mathbb{N})$ and consider the left and right shift operators on the standard basis. That is

$$T_L( (a_1, a_2, ... a_n , ... )) = (a_2, a_3, ... a_n,\ ... ) \\ T_R( (a_1, a_2, ... a_n , ... )) = (0, a_1,\ ... \ a_n,\ ... ) $$

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  • $\begingroup$ What if $T(e_n)=\sqrt(\frac{n+1}{n})e_n$ or some other number into $e_n$? $\endgroup$ – Ziya Sep 15 '18 at 3:43
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A bounded operator is completely determined by where it sends each $e_n$. Specifically, every element of $H$ has the form $\sum a_ne_n$ for some scalars $a_n$ (with $\sum |a_n|^2<\infty$), and then we must have $T(\sum a_ne_n)=\sum a_nT(e_n)$.

So, your task is to determine, in each of the three cases, whether the formula $$T\left(\sum a_ne_n\right)=\sum a_nT(e_n)$$ actually gives a well-defined bounded operator. This means that you need there exist a constant $C$ such that for all $(a_n)\in \ell^2$, $$\left\|\sum a_nT(e_n)\right\|\leq C\left(\sum |a_n|^2\right)^{1/2}$$ (and in particular, the infinite sum $\sum a_nT(e_n)$ needs to actually converge). Given the very simple definitions of $T(e_n)$ in each of your three cases, you should be able to write down a simple formula for $\left\|\sum a_nT(e_n)\right\|$ in each case which you can use to try to determine whether such a $C$ exists.

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  • $\begingroup$ I am still not getting anywhere from your answer. Could you please elaborate it for me? Actually I am self learning functional analysis. So I have no teacher to take help from. $\endgroup$ – Ziya Sep 14 '18 at 17:55
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    $\begingroup$ At least could you do it for $c$ part? $\endgroup$ – Ziya Sep 14 '18 at 17:56

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