# Range of operator $Mf(x) = f(x/2), \;\; x\in[0,1]$

Let $M: C([0,1]) \rightarrow C([0,1])$ be defined by $$Mf(x) = f(x/2), \;\; x\in[0,1]$$

Prove that the range of $I-M$ does not contain nonzero constant functions, but it contains all functions $C([0,1])$ that are differentiable (from the right) at 0 and satisfy $f(0) = 0$

My try: if $(I - M)f = const$ then the derivative $f'(0)$ would be infinite and hence the function not continuous. By this it also seems to follow that differentiability at 0 and f(0) = 0 must be necessary conditions for $g \in R(I-M)$ (R is the range). But how can I prove that we can reach all such functions?

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I was aware of this, but I did not know that this contradicted that $(I-M)f(x) = c$ directly – Johan Dec 26 '12 at 17:43

It is easy to see that $g(0)=0$ for all $g$ in the range of $I-M$. If $g$ is in the range and it is constant then there exists $f$ such that $f(x)-f(x/2)=g$ for all $x$. This implies that $f(x)=g+f(x/2)=2g+f(x/4)=\cdots$. Iterating this procedure we see that $f(x)$ is greater that any finite value and this is absurd (I am using that $f$ is continuous and that $f(0)=0$).
If $g$ is differentiable at $0$ then there exists $C$ such that $|g(x)|\leq Cx$ for all $x$. Now, define $f(x)=\sum g(\frac{x}{2^n})$. It follows that the series converges point-wise (and uniformly) and that $(I-M)f=g$.
You should put absolute values : $|g(x)| \leq Cx$ instead of just $g(x) \leq Cx$. – Ewan Delanoy Dec 26 '12 at 17:38
great! can you please expand how we get $|g(x) \leq Cx|$ and why we get uniformly convergence! – Johan Dec 26 '12 at 18:04
For small $x$ we have $|\frac{g(x)}{x}-g'(0)|\leq 1$, so that $g(x)\leq g'(0)+1$ and $g(x)\geq 1-g'(0)$. This gives you the bound for small $x$. For large $x$ it follows from the continuity. The convergence follows from the M-Test (en.wikipedia.org/wiki/Weierstrass_M-test) – Quimey Dec 26 '12 at 18:10