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Jochen Wengenroth
Jochen Wengenroth
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A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous identity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.

EDIT. You do not need a sledgehammer to show that $SOT$ differs from the operator norm topology: Fixing a non-zero element $y\in X$ we have an embedding $X^*\to B(X)$, $f\mapsto f\otimes y$ where $f\otimes y$ maps $x$ to $f(x)y$. The induced topologies on $X^*$ are the weak$^*$ topology and the dual norm topology. They differ because every continuous semi-norm of the weak$^*$ topology has a huge kernel.

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous identity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous identity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.

EDIT. You do not need a sledgehammer to show that $SOT$ differs from the operator norm topology: Fixing a non-zero element $y\in X$ we have an embedding $X^*\to B(X)$, $f\mapsto f\otimes y$ where $f\otimes y$ maps $x$ to $f(x)y$. The induced topologies on $X^*$ are the weak$^*$ topology and the dual norm topology. They differ because every continuous semi-norm of the weak$^*$ topology has a huge kernel.

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Jochen Wengenroth
Jochen Wengenroth
  • 16.5k
  • 2
  • 42
  • 82

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous idetityidentity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous idetity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous identity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.

Source Link
Jochen Wengenroth
Jochen Wengenroth
  • 16.5k
  • 2
  • 42
  • 82

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous idetity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.