Inequality on Hilbert spaces in order to prove the nonexpansivity of a mapping.

Question

I have an application $T\colon H\to H$ (where $H$ is a Hilbert space) such that $$(Tx-Ty,x-y)\leq \|x-y\|^2,\forall x,y\in H$$ where $(\cdot,\cdot)$ is the inner product of $H$ and $\|\cdot\|$ its associated norm.

I need to prove that $T$ is nonexpansive, i.e., $\|Tx-Ty\|\leq \|x-y\|,\forall x,y\in H$.

I have tried several things as the Cauchy-Schwarz inequality, the polarization identity, but I didn't goal the result.

Thanks in advance.

Answer 1

It is true if you assume that $T$ is self-adjoint (i.e. symmetric), meaning that $$ (Tx,y)=(x,Ty), \quad \text{for all}\,\, x,y\in H, \tag{1} $$ and assuming that $$ |(Tx,x)|\le \|x\|^2, \quad \text{for all}\,\, x\in H.\tag{2} $$ Note that your inequality holds even for $T=-2I$, and thus we NEED to assume these two additional things: $(1)$ and $(2)$.

So with the above assumptions, $$ (Tx+Ty,x+y)=(Tx,x)+(Ty,y)+2(Tx,y)\le \|x+y\|^2, $$ and $$ (Tx-Ty,x-y)=(Tx,x)+(Ty,y)-2(Tx,y)\ge -\|x-y\|^2, $$ thus $$ 4(Tx,y)\le \|x+y\|^2+\|x-y\|^2=2\big(\|x\|^2+\|y\|^2\big). $$ Finally, $$ \|Tx\|=\sup_{\|y\|=1}(Tx,y)\le \big(\|x\|^2+1)/2, $$ and therefore $$ \|T\|=\sup_{\|x\|=1}\|Tx\|\le 1, $$ which of course means that $$ \|Tx-Ty\|\le \|x-y\|, \quad \text{for all}\,\, x,y\in H. $$