Infimum of norm equals infimum of inner product for self-adjoint operators? Let $T$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$. It is known that
$$ \sup_{\|x\|=1} \|Tx\| = \sup_{\|x\|=1} |(Tx,x)|, $$ see for instance the answer given here.
Is there a way to prove that the same equality holds with supremum replaced by infimum on both sides? Using Cauchy-Schwarz we easily obtain one inequality, so it remains to prove that
$$ \inf_{\|x\|=1} \|Tx\| \le \inf_{\|x\|=1} |(Tx,x)|. $$
The standard strategy in the sup case is to use polarization identity for $|(Tx,y)|$, but it seems that this now eventually leads to the converse inequality.
 A: I would have liked to find something neater, but here is an argument.
Suppose first that $T$ is of the form
$$\tag1
T=\sum_j\lambda_jP_j, 
$$
where $\{P_j\}$ are pairwise orthogonal projections that add to the identity. Then
$$
\|Tx\|^2=\Big\|\sum_j\lambda_jP_jx\Big\|^2=\sum_j|\lambda_j|^2\,\|P_jx\|^2.
$$
It follows (by taking $x$ in the range of $P_j$ corresponding to $j$ with small $|\lambda_j|$) that
$$
\inf_{\|x\|=1}\|Tx\|=\inf\{|\lambda_j|:\ j\}. 
$$
Also,
$$
\langle Tx,x\rangle=\sum_j\lambda_j\|P_jx\|^2.
$$
So
$$
\inf_{\|x\|=1}|\langle Tx,x\rangle|=\inf\{|\lambda_j|: j\}=\inf_{\|x\|=1}\|Tx\|. 
$$
For arbitrary $T$, fix $\varepsilon>0$. Then there exists $T'$, of the form $(1)$, with $\|T-T'\|<\varepsilon$. As
$$
\big|\,\|Tx\|-\|T'x\|\,\big|\leq\|Tx-T'x\|\leq\|T-T'\|\,\|x|<\varepsilon\,\|x\|
$$
and
$$
|\langle Tx,x\rangle-\langle T'x,x\rangle|=|\langle (T-T')x,x\rangle|\leq\varepsilon \,\|x\|,
$$
we get that
$$
\inf_{\|x\|=1}|\langle Tx,x\rangle|<\varepsilon+\inf_{\|x\|=1}|\langle T'x,x\rangle|=\varepsilon+\inf_{\|x\|=1}\|T'x\|\leq2\varepsilon+\inf_{\|x\|=1}\|Tx\|.
$$
Similarly,
$$
\inf_{\|x\|=1}\|Tx\|\leq\varepsilon+\inf_{\|x\|=1}\|T'x\|=\varepsilon+\inf_{\|x\|=1}|\langle T'x,x\rangle|\leq2\varepsilon+\inf_{\|x\|=1}|\langle Tx,x\rangle|.
$$
Thus
$$
\big| \inf_{\|x\|=1}\|Tx\|-\inf_{\|x\|=1}|\langle Tx,x\rangle\big|\leq2\varepsilon. 
$$
As we are free to choose $\varepsilon$, this shows that
$$
\inf_{\|x\|=1}\|Tx\|=\inf_{\|x\|=1}|\langle Tx,x\rangle|. 
$$
