# Prove that a functional is convex

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

$$\Phi(x)=\frac{1}{2}(Tx,x)$$

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:

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

• If you are free to choose the path of proof and are stuck with the one you have chosen, then alternatively you may show that the function $t\mapsto \langle T(x+ty), x+ty \rangle$ is convex on $[0,1]$ for every $x,y$ (Why?). Since this is real function which is quite obviously smooth, you can calculate the second derivative and check whether it is positive. Commented Jun 20, 2015 at 11:56
• @Thomas I am not able to see why would this prove the statement above :/ Commented Jun 20, 2015 at 12:00

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 $f(x) = x^2$.

• Thank you! ok two questions: 1) Why the square root and 2) I do not know about the Jensen's inequality, which form did you use here, since at wiki I could not exactly understand the connection. Commented Jun 20, 2015 at 12:57
• @Mitscaype I added the Jensen step in there explicitly. It comes directly from the statement that $x^2$ is convex. Regarding why the square root. I needed it to apply the Cauchy Schwarz inequality in line three. Commented Jun 20, 2015 at 13:06
• Thank you! But I cannot see how can I write $(Tx,y)+(Ty,x)=(T^{1/2}x,T^{1/2}y)+(T^{1/2}y,T^{1/2}x)$? :/ Commented Jun 20, 2015 at 13:37
• @Mitscaype The square root is self-adjoint. Since $T=(T^{1/2})^2$, you can move one of the roots over to the RHS of the inner product. Commented Jun 20, 2015 at 13:40

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.

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.

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