Multiplication operator $M_f$ defined by $M_f g = f(x) g(x)$ is compact iff $f \equiv 0$

Question

Fix a continuous function $f: [0,1] \rightarrow \mathbb{R}$. Consider the multiplication operator $M_f: C^0 [0,1] \rightarrow C^0 [0,1]$ defined by $(M_f g)(x) = f(x)g(x)$ for all $x \in [0,1]$ and $g \in C^0([0,1])$. Show that $M_f$ is a compact operator if and only if $f \equiv 0$.

I'm having trouble with the forwards direction of the proof, where we assume $M_f$ is compact. I think I need to think of a bounded sequence $(g_n)$ whose image under $M_f$ cannot have any convergent subsequences unless $f = 0$, but I'm having trouble thinking of such a sequence.

Answer 1

Let $a \in (0,1)$ and construct a sequence of continuous functions $(g_n)$ with the properties $g_n(a)=1, g_n(x)=0$ if $x >a+\frac 1 n$ or $x <a-\frac 1 n$, $0\leq g(x)\leq 1$ for all $x$. (The graph is a triangle). Then $(g_n)$ us bounded in $C^{0}[0,1]$ so $(f(x)g_n(x))$ must have a convergent subsequence in $C^{0}[0,1]$. The limiting function is necessarily continuous. Note that the pointwise limit is $0$ for $x \neq a$. By continuity the limit has to be $0$ at all points. But $f(a)g_n(a)=f(a)$ so we must have $f(a)=0$. This is true for $0<a<1$ and, by continuity $f$ vanishes at the end points also.

Answer 2

To start, note that, if $M_f$ is compact, then $M_{\lambda f}$ is compact for any $\lambda \in \Bbb{R}$.

Suppose $f \neq 0$. Then either $f(x)$ must achieve a positive value or negative value somewhere. By replacing $f$ with $-f$ as necessary, assume that it achieves a positive value.

Continuity then implies that there is some non-trivial subinterval $[a, b]$ and some $\varepsilon > 0$ such that $$x \in [a, b] \implies f(x) > \varepsilon.$$ By replacing $f$ with $\frac{1}{\varepsilon} f$, we may assume without loss of generality that $\varepsilon = 1$.

The next WLOG step I wish to make is to show that $[a, b]$ can be assumed, WLOG, to be $[0, 1]$. For $g, h \in C[a, b]$, define $M'_h(g) = hg$. I claim that, if $M_f$ is compact, then $M'_{f|_{[a, b]}}$ is also compact. To see this, recall that the restriction map $g \mapsto g|_{[a, b]}$ is a non-expansive linear map from $C[0, 1]$ to $C[a, b]$. Then, given any bounded sequence $(g_n) \in C[a, b]$, then let $$g'_n \in C[0, 1] : x \mapsto \begin{cases} g_n(x) & \text{if } x \in [a, b] \\ g_n(a) & \text{if } x < a \\ g_n(b) & \text{if } x > b. \end{cases}$$ Then $\|g'_n\| = \|g_n\|$ and $g'_n|_{[a, b]} = g_n$. Then, $g'_n$ must have a Cauchy subsequence. Since the restriction map is non-expansive, the corresponding subsequence of $g_n$ is also Cauchy. Thus, if $M'_{f|_{[a, b]}}$ is not compact, then neither is $M_f$.

Let $\phi : C[a, b] \to C[0, 1] : g \mapsto \left(x \mapsto f\left(\frac{x - a}{b - a}\right)\right)$. Then $\phi$ is an isometric isomoprhism between the spaces, and given any $h \in C[a, b]$, $$M_{\phi(h)} \circ \phi = M'_h.$$ What this tells us is that if $M_{\phi(f|_{[a, b]})}$ is not compact, then neither is $M'_{f|_{[a, b]}}$, and hence neither is $M_f$. So, we can, without loss of generality, replace $f$ with $\phi(f|_{[a, b]})$, which is a function on $[0, 1]$ which is greater than $1$ on its entire domain! That is, WLOG, we may replace $[a, b]$ with $[0, 1]$.

Now, our last WLOG step! We can replace $f$ WLOG by the constant function $1$. Since $f(x) > 1$ for all $x$, it follows that $$\|M_f(g)\| = \sup_{x \in [0, 1]} f(x)g(x) \ge \sup_{x \in [0, 1]} 1g(x) = \|M_1(g)\|,$$ hence if $M_1$ is non-compact, then neither is $M_f$.

But, $M_1$ is simply the identity map on the infinite-dimensional space $C[0, 1]$. This implies $M_1$ is not compact. Thus, $M_f$ is not compact either.