Prove that a functional is convex

Question

Let $T$ be a self-adjoint bounded operator on a Hilbert space $H$. We define the functional $\Phi$ as:

\begin{equation} \Phi(x)=\frac{1}{2}(Tx,x) \end{equation}

My exercise says that $\Phi$ is convex if $T$ is strictly positive. I know that I have to prove that for $x,y \in H, t\in[0,1]$ it holds:

\begin{equation} \Phi(tx+(1-t)y)\leq t\Phi(x)+(1-t)\Phi(y) \end{equation} but until now I have managed to prove that: \begin{equation} \Phi(tx+(1-t)y)=t^2\Phi(x)+t(1-t)((Tx,y)+(Ty,x))+(1-t)^2\Phi(y) \end{equation} I know that I have to use the fact that $T$ is strictly positive, but I do not see how to do that..

Answer 1

To complete the proof you can observe that since $T$ is self-adjoint, positive, and bounded it has a square-root. Then \begin{eqnarray*} \Phi(tx+(1-t)y) &=& t^2\Phi(x)+t(1-t)((Tx,y)+(Ty,x))+(1-t)^2\Phi(y) \\ &=& t^2\Phi(x)+t(1-t)((T^{1/2}x,T^{1/2}y)+(T^{1/2}y,T^{1/2}x))+(1-t)^2\Phi(y) \\ &\leq& t^2\Phi(x)+2t(1-t)(T^{1/2}x,T^{1/2}x)(T^{1/2}y,T^{1/2}y)+(1-t)^2\Phi(y) \\ &=& t^2\Phi(x)+2t(1-t)\sqrt{\Phi(x)\Phi(y)}+(1-t)^2\Phi(y) \\ &=& \left(t\sqrt{\Phi(x)}+(1-t)\sqrt{\Phi(y)}\right)^2 \\ &=& t\Phi(x)+(1-t)\Phi(y) \\ \end{eqnarray*}

The last line follows from Jensen's inequality applied to $f(x) = x^2$.

Answer 2

An alternative is to use this identity for quadratic forms: $$ Q(\lambda x+(1-\lambda)y) = \lambda Q(x) + (1-\lambda) Q(y) - \lambda (1-\lambda) Q(x-y). $$ It's then pretty obvious that $\lambda (1-\lambda) Q(x-y)>0$ if $0<\lambda<1$ and $Q>0$, so the convexity is obvious.

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Now, you may be concerned about this identity applying here. $\Phi(x)=(Tx,x)$ is a quadratic form for real scalars: $$ \Phi(\lambda x) = (T(\lambda x,\lambda x) = \lvert \lambda \rvert^2 (Tx,x) = \lambda^2 (Tx,x) = \lambda^2 \Phi(x), $$ and $$ \Phi(x+y)-\Phi(x)-\Phi(y) = (T(x+y),x+y)-(Tx,x)-(Ty,y) = (Tx,y)+(Ty,x), $$ and $(T(\lambda x),y)+(Ty,\lambda x)=\bar{\lambda}(Tx,y)+\lambda(Ty,x)= \lambda((Tx,y)+(Ty,x))$. We are considering real $\lambda$, so this is all fine.

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The identity itself is quite straightforward to prove: if $Q(x+y)-Q(x)-Q(y)=2B(x,y)$ is the bilinear, $$ Q(\lambda x + (1-\lambda)y) = \lambda^2 Q(x)+(1-\lambda)^2 Q(y) + \lambda (1-\lambda) 2B(x,y) \\ = \lambda^2 Q(x)+(1-\lambda)^2 Q(y) + \lambda (1-\lambda) \left( Q(x)+Q(y)-Q(x-y) \right) \\ = \left( \lambda^2+\lambda(1-\lambda) \right)Q(x) + \left( (1-\lambda)^2 +\lambda (1-\lambda) \right)Q(y) -\lambda(1-\lambda)Q(x-y) \\ = \lambda Q(x) + (1-\lambda) Q(y) -\lambda(1-\lambda)Q(x-y). $$