Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $H$ be a real Hilbert space. The linear operator $A: H\rightarrow H$ is said to be positive definite iff there exists $M>0$ such that $$ {\rm (i)} \quad\langle Ax, x\rangle \geq M \|x\|^2, \quad \forall x\in H. $$ We have known that when $\text{dim}H<\infty$, the above inequality is equivalent to $$ {\rm (ii)}\quad \langle Ax, x\rangle > 0, \quad \forall x\in H\setminus\{0\}. $$ I would like to ask why people use (i) but do not use (ii) for the definition of positive definite operator in infinite dimensional space.

Thank you for all helping and comments.

share|cite|improve this question
Can you please give some examples or give some reference for the definition (i)?! – La Rias Jun 23 at 5:37
up vote 4 down vote accepted

If $V$ is finite dimensional, then you can either use i) or ii) for the definition of positive definiteness, although it seems i) is stronger. To see this, apply ii) to the unit ball in $V$, which is compact, hence there is $\delta>0$ such that \begin{equation} <Ax,x>\ge\delta \end{equation} for all $x$ on the unit ball, then linearity leads to i). Therefore the two definitions are equivalent for finite dimensional cases.

But i) is indeed stronger than ii) in infinite dimensional spaces.

According to my limited knowledge, the usefulness of positive definiteness lies in the fact that the bilinear form\begin{equation} (x,y)\mapsto <Ax,y> \end{equation} is bounded from below uniformly, as in i). (Of course it is bounded from above if you assume $A$ is a bounded linear operator.) This boundedness give very nice results, see the theorem by Lax-Milman, which is not true if you only assume ii), which allows the evil existence like \begin{equation} <Ax_n,x_n>\to 0. \end{equation}

One illustrating example comes from the existence of weak solutions to second order elliptic PDEs, which are of the form \begin{equation} Lu=-\sum_{i,j}\partial_j (a_{i,j}\partial_i u)+\sum_i \partial_i(b u)+cu, \end{equation} where the coefficients $a_{i,j}$, $b$ and $c$ are smooth functions.

Here we have the notion of ellipticity, which says $(a_{i,j}(x))$ (which is a finite dimensional matrix at each $x$) is positive definite, and uniform ellipticity, which says there is $M>0$ such that $<a_{i,j}(x)y,y>\ge M|y|^2$ for all $x$. The latter notion guarantees the existence of weak solutions. SO again you see bounded from below uniformly is important. (Here uniform in $x$).

share|cite|improve this answer
Thank you for your detailed instructions and interesting comments. – blindman Nov 1 '12 at 4:47

Another reason is the relation between spectrum and numerical range...

For bounded operators: $$A\in\mathcal{B}(\mathcal{H}):\quad\sigma(A)\subseteq\overline{\mathcal{W}(A)}$$

For normal operators: $$N^*N=NN^*:\quad\langle\sigma(N)\rangle=\overline{\mathcal{W}(N)}$$

The first case fails: $$\mathcal{W}(T)>0\nRightarrow\overline{\mathcal{W}(T)}>0$$

The second case works: $$\mathcal{W}(T)\geq\varepsilon>0\Rightarrow\overline{\mathcal{W}(T)}>0$$

That gives invertibility: $$0\notin\sigma(T)\implies T^{-1}\in\mathcal{B}(\mathcal{H})$$

But invertibility is important!

share|cite|improve this answer
For normal operators it should be the closed convex hull of the spectrum that equals to $\overline{\mathcal{W}(A)}$. – A.G. Oct 3 '15 at 18:16
Oh thanks for the hint. Corrected! – Alexander Frei Oct 4 '15 at 9:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.