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 [itex]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.

Try doing this using the same ideas here. You'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.

If you don'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.