Return to Answer

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le \|T\|$ and it is natural to ask if a $T$-independent constant $c > 0$ exists with $c\|T\| \le r(T)$, so that $r$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex (and $\dim H > 0$), the answer is yes and you can choose $c = \frac 12$.

If $H$ is real, the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Edit: According to

Martín, Miguel. A survey on the numerical index of a Banach space. III Congress on Banach Spaces (Jarandilla de la Vera, 1998). Extracta Math. 15 (2000), no. 2, 265--276. MR1823892

the proof for the complex case can be found on page 114 of

Halmos, Paul R. A Hilbert space problem book. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto, Ont.-London 1967 xvii+365 pp. MR0208368

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le \|T\|$ and it is natural to ask if a $T$-independent constant $c > 0$ exists with $c\|T\| \le r(T)$, so that $r$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex (and $\dim H > 0$), the answer is yes and you can choose $c = \frac 12$.

If $H$ is real (and $\dim H \ne 1$), the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Edit: According to

Martín, Miguel. A survey on the numerical index of a Banach space. III Congress on Banach Spaces (Jarandilla de la Vera, 1998). Extracta Math. 15 (2000), no. 2, 265--276. MR1823892

the proof for the complex case can be found on page 114 of

Halmos, Paul R. A Hilbert space problem book. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto, Ont.-London 1967 xvii+365 pp. MR0208368

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le T$ and it is natural to ask if a $T$-independent constant $c > 0$ exists with $c\|T\| \le r(T)$, so that $r$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex (and $\dim H > 0$), the answer is yes and you can choose $c = \frac 12$.

If $H$ is real, the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Edit: According to

Martín, Miguel. A survey on the numerical index of a Banach space. III Congress on Banach Spaces (Jarandilla de la Vera, 1998). Extracta Math. 15 (2000), no. 2, 265--276. MR1823892

the proof for the complex case can be found on page 114 of

Halmos, Paul R. A Hilbert space problem book. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto, Ont.-London 1967 xvii+365 pp. MR0208368

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le \|T\|$ and it is natural to ask if a $T$-independent constant $c > 0$ exists with $c\|T\| \le r(T)$, so that $r$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex (and $\dim H > 0$), the answer is yes and you can choose $c = \frac 12$.

If $H$ is real, the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Edit: According to

Martín, Miguel. A survey on the numerical index of a Banach space. III Congress on Banach Spaces (Jarandilla de la Vera, 1998). Extracta Math. 15 (2000), no. 2, 265--276. MR1823892

the proof for the complex case can be found on page 114 of

Halmos, Paul R. A Hilbert space problem book. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto, Ont.-London 1967 xvii+365 pp. MR0208368

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le T$ and it is natural to ask if a constant $c > 0$ exists such that $c\|T\| \le r(T)$, so that $r(T)$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex, the answer is yes and you can choose $c = \frac 12$.

If $H$ is real, the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le T$ and it is natural to ask if a $T$-independent constant $c > 0$ exists with $c\|T\| \le r(T)$, so that $r$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex (and $\dim H > 0$), the answer is yes and you can choose $c = \frac 12$.

If $H$ is real, the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Edit: According to

Martín, Miguel. A survey on the numerical index of a Banach space. III Congress on Banach Spaces (Jarandilla de la Vera, 1998). Extracta Math. 15 (2000), no. 2, 265--276. MR1823892

the proof for the complex case can be found on page 114 of

Halmos, Paul R. A Hilbert space problem book. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto, Ont.-London 1967 xvii+365 pp. MR0208368