# Continuous linear image of closed, bounded, and convex set of a Hilbert Space is compact

Is my proof of this proposition correct ? And is this proposition well known?

Proposition: Let $C$ be a closed, bounded, and convex set in a separable Hilbert space $H$. Let $L : H \to \mathbb{R}^n$ be a continuous linear transformation. Then $L(C)$ is compact in $\mathbb{R}^n$.

Proof: Since $H$ is a Banach space, and $C$ is closed and convex, $C$ is weakly closed (Mazur's Theorem in Lang's Real Analysis). Since $H$ is a reflexive Banach space, closed balls are weakly compact (Kakutani's Theorem). Since $C$ is bounded it is contained in a weakly compact ball, and since $C$ is weakly closed, $C$ is weakly compact.
Since $L$ is continuous in the strong topologies, it is continuous in the weak topologies (this fact seems to be well known). So $L(C)$ is weakly compact in $\mathbb{R}^n$, and since the weak and strong topologies on $\mathbb{R}^n$ are the same, $L(C)$ is compact in $\mathbb{R}^n$. $\square$

Convexity is essential. Otherwise I have a counterexample.

This is not a homework problem. The motivation comes from the physics of color. In my case $H$ is $L^2([380,780])$ where 380 and 780 are wavelengths of light in nanometers. $C$ is the subset of functions that take values in $[0,1]$ and each such function represents the spectral reflectance (or spectral transmittance) of a material over this interval of wavelengths. $\mathbb{R}^n$ is the 3D space of CIEXYZ tristimulus coordinates. $L$ is the linear mapping from the reflectance of the material to CIEXYZ, for a fixed spectral illuminant. $L(C)$ is the set of all possible material colors for the given illuminant.

Thank you.

• Yes, your proof is correct. I can't cite a book or anything, but I'm pretty sure it's well enough known, people have investigated (weakly) compact convex sets a lot. – Daniel Fischer Aug 24 '14 at 21:57
• Just for interest: Whats your counterexample? – Daniel Aug 25 '14 at 2:14
• @Daniel: An orthonormal basis would serve as a counterexample. Let $L:H\to\mathbb R$ be defined by $L(x)=\langle x,y\rangle$, with $y$ chosen to have nonzero inner product with each element of the chosen orthonormal basis. – Jonas Meyer Aug 25 '14 at 5:04
• @Daniel: The counterexample of Jonas is the same as mine. If $C$ is an orthonormal basis, then $L(C)$ has zero as a limit point, but does not contain zero. – Glenn Davis Aug 25 '14 at 6:56

More to the point, here is another presentation of essentially the same proof. Take any sequence in $L(C)$ and write it as $(L(x_n))$. Using the aforementioned weak* compactness (and reflexivity), choose a weakly convergent subsequence of $(x_n)$, denoted $(x_{n_k})$. The weak convergence implies convergence of $L(x_{n_k})$, since $L$ is just a finite number of linear functionals.
Also, the weak limit of $x_{n_k}$ lies in $C$, for otherwise we'd be able to separate it from $C$ by a hyperplane, contradicting weak convergence. Hence, the limit of $L(x_{n_k})$ lies in $L(C)$, proving compactness.