# Does $\|f(T^*T)T^*\|_\infty = \|f(T^*T)(T^*T)^{\frac{1}{2}}\|_\infty$?

If $T:X\to Y$ is a compact operator and $X,Y$ are some Hilbert spaces, can we say that $\|f(T^*T)T^*\|_\infty = \|f(T^*T)(T^*T)^{\frac{1}{2}}\|_\infty$, where $T^*$ is its adjoint and $f$ some continuous function.

Update: The only thing that I can show is that $\|T\|_\infty=\|T^*T\|_\infty^{\frac{1}{2}}=\|(T^*T)^{1/2}\|_\infty$

I can also show that $\|f(T^*T)T^*\|^2_\infty = \|Tf(T^*T)^2T^*\|_\infty$, but I can't see how this could help.

For any bounded linear operator $A : X\rightarrow Y$ between complex Hilbert spaces $X$, $Y$, the adjoint $A^{\star}$ is bounded from $Y$ to $X$ and $$\|A\|_{\mathcal{L}(X,Y)}=\|A^{\star}\|_{\mathcal{L}(Y,X)}$$ But it is also true that $$\|A\|_{\mathcal{L}(X,Y)}^{2}=\sup_{\|x\|_{X}=1}\|Ax\|_{Y}^{2} =\sup_{\|x\|_{X}=1}(A^{\star}Ax,x)_{X}=\|A^{\star}A\|_{\mathcal{L}(X,X)}.$$ In your case, you have $A=BT^{\star}$ where $B\in\mathcal{L}(X,X)$ and $T\in\mathcal{L}(X,Y)$. So $A : Y \rightarrow X$ in this case. This gives \begin{align} \|A\|_{\mathcal{L}(Y,X)}^{2} & =\|A^{\star}\|_{\mathcal{L}(X,Y)}^{2}\\ & =\|AA^{\star}\|_{\mathcal{L}(X,X)} \\ & =\|BT^{\star}TB^{\star}\|_{\mathcal{L}(X,X)}\\ & =\|B(T^{\star}T)^{1/2}(T^{\star}T)^{1/2}B^{\star}\|_{\mathcal{L}(X,X)} \\ & = \|B(T^{\star}T)^{1/2}\|_{\mathcal{L}(X,X)}^{2} \end{align} Your $B=f(T^{\star}T)$, which gives the final result: $$\|f(T^{\star}T)T^{\star}\|_{\mathcal{L}(X,X)}=\|f(T^{\star}T)(T^{\star}T)^{1/2}\|_{\mathcal{L}(X,X)}$$