Let $H$ be a Hilbert space. For a bounded linear operator $T:H\to H$ we write $T\geq 0$ to mean that $T$ is self-adjoint and that $\langle Tx, x\rangle \geq 0$ for all $x\in H$. For two bounded linear operators $S$ and $T$ we will write $S\geq T$ to mean $S-T\geq 0$.

Suppose $T_1, T_2, T_3, \ldots$ is a sequence of self-adjoint operators on $H$ such that

1) $T_{n+1}\geq T_n$ for all $n$.

2) There is a self adjoint operator $T'$ such that $T'\geq T_n$ for all $n$.

Problem. Show that there is a self adjoint operator $T$ such that $T_nx\to Tx$ (convergence in norm) for all $x\in H$.

It is clear that $\langle T_nx, x\rangle$ is a bounded and increasing sequence of real numbers and hence has a limit. Therefore so does $\langle T_n(x+y), x+y\rangle$. From here we can conclude that $\langle T_nx, y\rangle$ is a Cauchy sequence for each $x$ and $y$ in $H$.

But what I really need to show is that $\{T_nx\}$ is a Cauchy sequence and I am unable to do that.


1 Answer 1


The sequence $(T_{n}x,x)$ is convergent: $(T'x,x)\geq(T_{n+1}x,x)\geq(T_{n}x,x)$ for each $n=1,2,...$

Then sequence $(T_{n}x,y)$ is also convergent by means of polarization.

We can define $(Tx,y)=\lim_{n}(T_{n}x,y)$. It is easy to see that $T$ is self-adjoint because of those $T_{n}$.

On the other hand, by Uniform Boundedness Principle, one can show that $\|T_{n}\|\leq C$ and hence $T$ is also bounded.

Indeed, consider the sesquilinear form $U(x,y)=(T_{n}x,y)$, then we have \begin{align*} \|T_{n}x\|^{2}&=U(x,T_{n}x)\\ &\leq U(x,x)^{1/2}U(T_{n}x,T_{n}x)^{1/2}\\ &=(T_{n}x,x)^{1/2}(T_{n}^{2}x,T_{n}x)^{1/2}\\ &\leq(T_{n}x,x)^{1/2}(T'(T_{n}x),T_{n}x)^{1/2}\\ &\leq(T_{n}x,x)^{1/2}\|T'\|^{1/2}\|T_{n}x\|, \end{align*} so $\|T_{n}x\|\leq(T_{n}x,x)^{1/2}\|T'\|^{1/2}\leq(T'x,x)^{1/2}\|T'\|^{1/2}$, so for each $x$, $\|T_{n}x\|$ is bounded, now we apply Uniform Boundedness Principle.

Now consider the sesquilinear form $S(x,y)=((T-T_{n})(x),y)$.

We have \begin{align*} \|(T-T_{n})x\|^{2}&=S(x,(T-T_{n})x)\\ &\leq S(x,x)^{1/2}S((T-T_{n})x,(T-T_{n})(x))^{1/2}\\ &=((T-T_{n})(x),x)^{1/2}((T-T_{n}x)^{2},(T-T_{n})(x))\\ &\leq((T-T_{n})(x),x)^{1/2}\|(T-T_{n})(x)\|^{1/4}\|(T-T_{n})^{2}(x)\|^{1/4}\\ &\leq C'((T-T_{n})(x),x)^{1/2}\|x\|^{1/2}\\ &\rightarrow 0. \end{align*}


The boundedness of $\|T_{n}\|$ can be proved without appealing to $T'$.

Indeed, the sequence $(T_{n}x,y)$ is bounded by applying polarization to the boundedness of $(T_{n}x,x)$.

Now we consider $S_{n}(y)=(y,T_{n}x)$, for each fixed $x$, Uniform Boundedness Theorem gives $\|S_{n}\|\leq M_{x}$, then $|(T_{n}x,y)|\leq M_{x}$ for any $y$ with $\|y\|\leq 1$. We set $y=T_{n}x/\|T_{n}x\|$ to get $\|T_{n}x\|\leq M_{x}$, once again Uniform Boundedness Theorem gives $\sup_{n}\|T_{n}\|<\infty$.

  • $\begingroup$ I am unable to see the first inequality in the last part. If we write $L$ in place of $T-T_n$, then we have $S(x, y) = (Lx, y)$. So the inequality becomes $\|Lx\|^2 \leq \sqrt{(Lx, x)}\sqrt{(L^2x, Lx)}$. I am unable to see why this should be the case. Can you please help. Thanks. $\endgroup$ Nov 2, 2019 at 16:32
  • 1
    $\begingroup$ The point is to use the Cauchy-Shwarz to the sesquilinear form: $S(x,y)\leq S(x,x)^{1/2}S(y,y)^{1/2}$. $\endgroup$
    – user284331
    Nov 2, 2019 at 16:39
  • $\begingroup$ When using the Cauchy-Schwarz for $U$, do we need to assume that $T_n$ is positive semi-definite, or just the self-adjointness of $T_n$ suffices? I ask this because $\langle T_nx, x\rangle$ might be negative and therefore its square root will cause problems. $\endgroup$ Nov 5, 2019 at 10:47
  • $\begingroup$ Positivity is needed, I almost forgot, but luckily in this case they are. $\endgroup$
    – user284331
    Nov 5, 2019 at 14:29

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