# Limit of an increasing sequence of self-adjoint operators

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.

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*}

Edit:

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 $$\sup_{n}\|T_{n}\|<\infty$$.

• 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. Nov 2, 2019 at 16:32
• The point is to use the Cauchy-Shwarz to the sesquilinear form: $S(x,y)\leq S(x,x)^{1/2}S(y,y)^{1/2}$. Nov 2, 2019 at 16:39
• 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. Nov 5, 2019 at 10:47
• Positivity is needed, I almost forgot, but luckily in this case they are. Nov 5, 2019 at 14:29