I am wondering

if there is any condition one can apply (e.g. uniform boundedness?) that ensure the adjoint of a net of SOT-continuous elements is again SOT-continuous?

My major question is

$\{v_t\}$ is a SOT continuous path of bounded operators, P is compact operator, then when is $\{v_tPv_t^*\}$ a norm continuous path of operators?

I got some partial answer in my previous question: Question about SOT and compact operators which shows me that $\{v_tP\}$ is norm-continuous, but I am not sure how to apply that to solve my major question. I was hoping the first question I raised in the current post may help answer it. Thank you!

  • $\begingroup$ You've found a mistake in my proof, so I've deleted. What a pity $\endgroup$
    – Norbert
    Oct 21, 2013 at 17:12
  • $\begingroup$ :( I am sorry. Do you have any further suggestion for me? Thanks anyway for your enthusiastic help! – you have reminded me the proof of uniform boundedness of a SOT-convergent sequence, and taught me the TeX code for $\Vert$ :) $\endgroup$ Oct 22, 2013 at 5:16
  • $\begingroup$ See new version of my proof. Hope this time it is ok. $\endgroup$
    – Norbert
    Oct 22, 2013 at 8:36

1 Answer 1


Preliminaries. Let $H$ be a Hilbert space. Consider bilinear operator $$ \bigcirc: H\times H^{cc}\to\mathcal{F}(H):(x,y)\mapsto (z\mapsto \langle z,y\rangle x) $$ where $H^{cc}$ is a complex conjgate Hilbert space and $\mathcal{F}(H)$ is a normed space of all finite rank operators with operator norm. One can check that $\bigcirc$ is bounded bilinear operator of norm one and its image is all rank one operators. For a given $x,y\in H$ this is straightforward to check that $$ A(x\bigcirc y)B=A(x)\bigcirc B^*(y)\tag{1} $$

Lemma Let $H$ be a Hilbert space and $(T_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $T\in\mathcal{B}(H)$. Then, $(T_n:n\in\mathbb{N})$ is norm bounded in $\mathcal{B}(H)$ by some constant $c_T$.

Proof. Since $(T_n:n\in\mathbb{N})$ strongly converges to $T$, then for all $x\in X$ the sequence $(T_n(x):n\in\mathbb{N})$ is norm bounded in $H$. By Banach-Steinhaus theorem the sequence $(T_n:n\in\mathbb{N})$ is norm bounded in $\mathcal{B}(H)$ by some constant $c_T>0$

Lemma. Let $H$ be a Hilbert space $A\in\mathcal{K}(H)$ and $(T_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $T\in\mathcal{B}(H)$, $(S_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $S\in\mathcal{B}(H)$ then so $(T_nAS_n^*:n\in\mathbb{N})$ converges to $TAS^*$ in the norm topology of $\mathcal{B}(H)$.

Proof. Let $x,y\in H$. Since $(T_n:n\in\mathbb{N})$ and $(S_n:n\in\mathbb{N})$ converges to $T$ and $S$ respectively in the strong operator topology, then $(T_n(x):n\in\mathbb{N})$ and $(S_n(y):n\in\mathbb{N})$ converge to $T(x)$ and $S(y)$ respectively in the norm topology of $H$. By properties of bilinear operator $\bigcirc$ we conclude that $(T_n(x)\bigcirc S_n(y):n\in\mathbb{N})$ converges to $T(x)\bigcirc S(y)$ in the norm topology of $\mathcal{B}(H)$. Therefore from $(1)$ we get that $$ \lim\limits_{n\to\infty} T_n(x\bigcirc y) S_n^*=T(x\bigcirc y)S^*\tag{2} $$ in the norm topology of $\mathcal{B}(H)$. Now take arbitrary $F\in\mathcal{F}(H)$. Since every finite rank operator is a sum of rank one operators, then there exists $(x_1,\ldots,x_k)\subset H$ and $(y_1,\ldots,y_k)\subset H$ such that $F=\sum_{i=1}^k x_k\bigcirc y_k$. So using $(2)$ we get $$ \lim\limits_{n\to\infty} T_nF S_n^* =\sum_{i=1}^k\lim\limits_{n\to\infty} T_n(x_k\bigcirc y_k) S_n^* =\sum_{i=1}^k T(x_k\bigcirc y_k)S^* =TFS^*\tag{3} $$ in the norm topology of $\mathcal{B}(H)$.

Finally, consider compact operator $A\in\mathcal{K}(H)$. Since $H$ is a Hilbert space, then $A$ is limit of finite rank operators in the norm topology of $\mathcal{B}(H)$. Fix $\varepsilon>0$. From the previous note we have that there exist $F\in\mathcal{F}(H)$ and $D\in\mathcal{K}(H)$ such that $A=F+D$ and $\Vert D\Vert<\varepsilon$. Then using previous lemma we get $$ \Vert T_n DS_n^*-TDS^*\Vert \leq\Vert T_n DS_n^*\Vert+\Vert TDS^*\Vert \leq\Vert T_n\Vert \Vert D\Vert \Vert S_n^*\Vert+\Vert T\Vert\Vert D\Vert \Vert S^*\Vert \leq c_Tc_S \varepsilon +\Vert T\Vert\Vert S\Vert \varepsilon $$ $$ 0\leq\Vert T_n AS_n^*-T AS^*\Vert \leq \Vert T_n FS_n^*-T FS^*\Vert+\Vert T_n DS_n^*-T DS^*\Vert\\ \leq \Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon $$

Now we take $\limsup$ when $n\to\infty$ and use $(3)$ to get $$ 0\leq\limsup\limits_{n\to\infty}\Vert T_n AS_n^*-T AS^*\Vert \leq \limsup\limits_{n\to\infty}\Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon\\ \leq \lim\limits_{n\to\infty}\Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon \leq (c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon $$ Since $\varepsilon>0$ is arbitrary from previous inequalities we conclude that $\lim\limits_{n\to\infty}\Vert T_n AS_n^*-T AS^*\Vert$ exists and equals to $0$. Hence we proved that $$ \lim\limits_{n\to\infty} T_n AS_n^*=T AS^* $$ in the norm topology of $\mathcal{B}(H)$.

Proposition Let $H$ be a Hilbert space and $P\in\mathcal{K}(H)$. Assume $\gamma:[0,1]\to\mathcal{B}(H):t\mapsto v_tpv_t^*$ is a path continuous in strong operator topology, then it is continuous in norm topology.

Proof Since $[0,1]$ is a first-countable topological space it is enough to show sequential continuity. For a given sequence $(t_n:n\in\mathbb{N})\subset[0,1]$ convrgent to $t\in[0,1]$ define $S_n=T_n=v_{t_n}$ and apply previous lemma to get that $v_{t_n}Pv_{t_n}^*$ converges to $v_tPv_t^*$ in norm. Hence $\gamma$ is norm continuous path.

  • $\begingroup$ I tried approaching through the following argument: $\sup_{x\in B_X}\Vert(AB-CD)(x)\Vert\leq \max\{\sup_{y\in S}\Vert(A-C)B(y)\Vert, \sup_{y\in S}\Vert(A-C)D(y)\Vert\}$ but it turned out to be false if we set $A=C$, $B=-D$... Any suggestion? $\endgroup$ Oct 21, 2013 at 14:46
  • $\begingroup$ Now you can delete this two comments since they are irrelevant $\endgroup$
    – Norbert
    Oct 22, 2013 at 8:37
  • $\begingroup$ Thank you so much Norbert! I believe this is a correct proof. Just a question: why did you choose $H^{cc}$ on line 2? Would anything go wrong if I have $H \times H$ instead and define the map by $z\mapsto <y,z>x$? $\endgroup$ Oct 24, 2013 at 6:17
  • 1
    $\begingroup$ In this case you will get anti-linear rank one operators instead of linear operators $\endgroup$
    – Norbert
    Oct 24, 2013 at 6:21
  • $\begingroup$ Ah yes. Thank you so much! $\endgroup$ Oct 24, 2013 at 18:49

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