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Let $T: X\to Y$ be a bounded linear operator between Banach spaces.Assume that there exists c$\in R$ and a compact operator $K: X\to Z$ such that

$$||x||_X\leq c||Tx||_Y+||Kx||_Z\quad \forall x\in X$$

Prove that

a)Ker(T) ha finite dimension

b) if $(x_n)$ is a sequence in X with bounded image and $||x_n||_X\to \infty$ then the sequence $x_n/||x_n||_X$ has a subsequence that converges to an element of Ker(T).

c) T has closed image

Prove that if $T\in L(X,Y)$ has a finite dimension kernel and closed image there exists c and K like before.

I would appreciate any idea on the solution. Thank you in advance.

user62138
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1 Answers1

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a) Suppose that $x\in\ker T$. Then $\|x\|_X\leq\|Kx\|_Z$. As $K$ is compact, this can only happen on a finite-dimensional subspace (otherwise, you would be able to put a ball in the image of $K$).

b) The assumption is that $\|Tx_n\|_Y\leq k$ for all $n$. Let $y_n=x_n/\|x_n\|_X$. Then $$ \|Ty_n\|_Y=\frac{\|Tx_n\|_Y}{\|x_n\|_X}\leq\frac{k}{\|x_n\|_X}\to0. $$ The compacity of $K$ allows us to choose a subsequence $y_{n_k}$ with $\{Ky_{n_k}\}$ convergent; in particular, it is Cauchy. As $\|y_{n_k}-y_{n_j}\|_X\leq c\|Ty_{n_k}-Ty_{n_j}\|_Y+\|Ky_{n_k}-Ky_{n_j}\|_Z$, we deduce that $\{y_{n_k}\}$ is Cauchy. As $X$ is Banach, there exists $y\in X$ with $y_{n_k}\to y$. Then, as $T$ is bounded, $Ty=\lim_kTy_{n_k}=0$, i.e. $y\in\ker T$.

c) The inequality $$ \|x\|_X\leq c\|Tx\|_Y+\|Kx\|_Z $$ will still hold if we replace $c$ with a smaller positive number. So we can assume that $c\leq\frac1{2\|T\|}$. Then $$ \|x\|_X\leq c\|Tx\|_Y+\|Kx\|_Z\leq c\|T\|\,\|x\|_X+\|Kx\|_z\leq\frac12\,\|x\|_X+\|Kx\|_Z, $$ and so $$ \|x\|_X\leq 2\|Kx\|_Z $$ for all $x\in X'$. Let $X_0$ be the closed span of $\{x\in X':\ Kx\ne0\}$. The compacity of $K$ forces $X_0$ to be finite dimensional; it is then complemented in $X'$, and this complement $X_1$ is contained in the kernel of $K$. So $X=\ker T\oplus X_0\oplus X_1$, with the first two finite-dimensional. So it is enough to check that $T(X_1)$ is closed. Let $\{Tx_n\}$ be a Cauchy sequence with $\{x_n\}\subset X_1$. Then $Kx_n=0$ for all $n$ and $\|x_{n_k}-x_{n_j}\|_X\leq c\|Tx_{n_k}-Tx_{n_j}\|_Y$, so $\{x_{n_k}\}$ is Cauchy. As $X_1$ is Banach, there exists $x\in X_1$ with $x_{n_k}\to x$, and then $\lim Tx_n=Tx$.

d) As $\ker T$ is finite-dimensional, it is complemented in $X$. Let $K$ be the projection onto $\ker T$, i.e. $K$ is the identity on $\ker T$ and zero on its complement $X'$. Now, for any $x\in X$ we have $x=x_1+x_2$, with $x_1\in\ker T$ and $x_2\in X'$. $$ \|x\|_X=\|x_1+x_2\|_X\leq\|x_1\|_X+\|x_2\|_X=\|Kx_1\|_X+\|x_2\|_X $$ Now, as $X'$ is closed, it is a Banach space, and the restriction $T:X'\to \text{im}\, T$ is bounded, one-to-one, and onto. So it has a bounded inverse; in particular, $T$ is bounded below on $X'$. So we have $$ \|x\|_X\leq\|Kx_1\|_X+\|x_2\|_X=\|Kx\|_X+\|x_2\|_X\leq\|Kx\|_X+c\|Tx_2\|_Y=\|Kx\|_X+\|Tx\|_Y. $$

Martin Argerami
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  • Could you clarify the argument you have provided for part c)? In particular, I'm not sure how you conclude that ${Tx_n}$ being Cauchy implies that ${x_n}$ is bounded. I think this is how you mean to show that ${Kx_{n_k}}$ is convergent? Thanks in advance. – JZS Dec 05 '13 at 23:47
  • You are totally right. The argument in c) was wrong. Look at the new version. – Martin Argerami Dec 06 '13 at 01:01
  • Sorry for so many comments, but in your new argument for c), you say that $T \colon X' \to T(X) $ is bounded, one-to-one, and onto . Thus, because $X'$ is Banach, the restriction has a bounded inverse. But (according to the open mapping theorem), don't we also need to know that $T(X)$ is Banach in order to make the claim that the inverse is bounded? – JZS Dec 06 '13 at 02:47
  • You are right again. Sorry about that. I will revert to the previous version, and I'll think about it tomorrow. – Martin Argerami Dec 06 '13 at 04:25
  • Thanks! I am only asking so many questions because I am working on a similar problem. That is, I am allowed to assume the inequality $||x||_X \le c||Tx||_Y + ||Kx||_Z$, and I need to show that $T(X)$ is closed. But I am also allowed to assume $T$ is injective. It's just difficult for me to show that ${Tx_n}$ being Cauchy implies ${x_n}$ is Cauchy. This seems to be the key point of the argument. – JZS Dec 06 '13 at 13:29
  • I'm trying a new idea in c). Please take a look. Note that the argument cannot be very direct, as it is easy to construct a compact operator $T$ and sequence $x_n$ with ${Tx_n}$ Cauchy and $|x_n|\to\infty$. – Martin Argerami Dec 06 '13 at 15:16