Is the hyperplane of equation [f=c] closed if and only if f is continuous?

[quasar987]

[SOLVED] hyperplanes and continuity

Homework Statement

Let X be a real normed linear space, f a linear functional on X and c a real constant. The set f^{-1}(c) is called the hyperplane of equation [f=c] and supposedly, the hyperplane of equation [f=c] is closed if and only if f is continuous.

Is this obvious? I don't see it.

The Attempt at a Solution

The part <== is obvious: f is continuous so the preimage of a closed set is a closed set. Therfor, since {c} is a closet set, so is the hyperplane f^{-1}(c).

What about the other direction?

 

[morphism]

Consider the natural mapping \pi : X \to X/f^{-1}(c). This is continuous (in fact contractive), surjective and linear. Let f^*:X/f^{-1}(c) \to \mathbb{R} be the induced linear embedding from f. Note that f^* \circ \pi is a linear functional on X whose kernel is f^{-1}(c)[/tex]. Conversely, the kernel of any linear functional is obviously a hyperplane. So it suffices to prove that f is continuous iff its kernel is closed.<br /> <br /> Try doing this using the same ideas here. You&#039;ll find it useful to keep in mind that if two linear functionals have the same kernel, then one is a scalar multiple of the other.<br /> <br /> If you don&#039;t like playing with quotient spaces, you can try showing that f is continuous at zero / bounded about zero.

 

[HallsofIvy]

Whether this is obvious or how you would prove it depends on what definitions you have to use. A standard definition for "continuous" is that a function f is continuous if and only if f^-1(U) is open whenever U is open. From that it follows that f^-1(V) is closed whenever V is closed (and sometimes that is used as the definition of "continuous"). That, together with the fact that a singleton set, such as {c} is closed in a Hilbert Space, does make this statement obvious!

 

[morphism]

I don't think it's that simple Halls! That f^-1({c}) is closed if f is continuous is certainly obvious as you say, but going the other direction is definitely harder.

 

[quasar987]

To conclude that f is continuous, it must be that all closed sets have closed preimage, not just a singleton {c} !

 

[HallsofIvy]

You're right. I misread that completely.

 

[quasar987]

My professor said "the hyper plane is close, so its complement is open. Chose a point in said complement. Then there is a ball centered on it a completely contained in said complement. Show that f is continuous at a.

Anyone sees how this argument works in the details?

 

[quasar987]

I asked her for details today.. it goes like this.

Call A the complement of the hyperplane of equation [f=c]. Since A is open, there exist an a in A and a ball of radius r centered on a and entirely contained in A. We can assume without loss of generality that f(a)<c. It must be also that all elements y in the ball satisfies f(y)<c (because if f(y)>c, by convexity, it must be that the ball intersects the hyperplane). This condition can be written as "for all ||z||<1, f(a+rz)<c". So by linearity, f(z)<(c-f(a))/r. Taking the sup over z, this gives the continuity of f.