A formulation of continuity for bilinear forms

[quasar987]

[SOLVED] A formulation of continuity for bilinear forms

Homework Statement

My HW assignment read "Let H be a real Hilbert space and a: H x H-->R be a coninuous coersive bilinear form (i.e.
(i) a is linear in both arguments
(ii) There exists M>0 such that |a(x,y)|<M||x|| ||y||
(iii) there exists B such tthat a(x,x)>a||x||^2"

So apparently, condition (ii) is the statement about continuity. But I fail to see how this statement is equivalent to "a is continuous".

I see how (ii) here implies continuous, but not the opposite.

The Attempt at a Solution

Let z_n = (x_n,y_n)-->0. Then x_n-->0 and y_n-->0. So |a(x,y)|<M||x|| ||y|| implies a(x,y)-->0. a is thus continuous at 0, so it is so everywhere, being linear.

 

[Dick]

I'm not quite sure what the question is. It seems to have been omitted. But just looking at your argument, z_n=(x_n,y_n)->0 doesn't imply x_n->0 or y_n->0. Does it?

 

[quasar987]

Well, I'm making use of the fact that if (M,d) is metric space, then the product topology on M x M is generated by the metric

D((x1,y1),(x2,y2))=[d(x1,x2)² + d(y1,y2)²]^½

I conclude that the norm on H x H is

||(x,y)|| = [||x||² + ||y||²]^½

And now if (x_n,y_n)-->0, this means that

[||x_n||² + ||y_n||²]^½ --> 0,

which can only happen if x_n-->0 and y_n-->0.

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You're right, I have not actually typed the question entirely. It is because I was confused by the fact that they seem to imply that condition (ii) is equivalent to continuity. While (ii) implies continuity, does continuity implies (ii)?

 

[Hurkyl]

Well, there is the theorem that a linear functional on a Hilbert space is continuous if and only if it's bounded...

 

[quasar987]

You mean bounded in the unit ball?

In this case, you're right, it works. Because ||(x,y)|| = [||x||² + ||y||²]^½ <1 ==> ||x||, ||y||<1 and (ii) implies|a(x,y)|<M for all (x,y) in H x H such that ||(x,y)||<1.