Showing that the uniform boundedness principle implies the meagerness of differentiable functions I'm trying to follow a proof that the set (call the set $M$) of functions that are differentiable at a fixed $a \in [0,1]$ is meager in the set of continuous functions (call this set $X$) from $[0,1]$ to $\mathbb{C}$.
The proof goes something like this:
Take a sequence $(f_n)$ of functionals on $X$ given by $f_n (\phi) = n(\phi(a + \frac{1}{n}) - \phi (a))$.
We can show that $|| f_n || = 2n$ by choosing $\phi$ such that $\phi(a + \frac{1}{n}) = 1$ and  $\phi(a) = -1$ -- or at least I think this is legitimate, because to calculate the norm of $f_n$ we need the supremum of $|f_n (\phi)|$ over all functions $\phi$ which achieve an absolute value of at most $1$, and it's clear that the construction we used will maximize $f_n (\phi)$ while sticking to that definition.
Now if we define $M_0 = \{\phi \in X : \sup_{n\in\mathbb{N}} |f_n (\phi)| < \infty \}$, apparently $M_0$ is meager in $X$, i.e. it's the union of countably many sets whose closures have no interior points. This is the step I don't understand. How does the uniform boundedness principle play a role in establishing this fact? I imagine the union (to show meagerness) is going to be $\{\phi \in X : \sup_{n\in\mathbb{N}} |f_n (\phi)| < 1 \} \cup \{\phi \in X : 1 \leq \sup_{n\in\mathbb{N}} |f_n (\phi)| < 2 \} \cup \cdots$, but I don't see anything about the interiors of the closures of those sets.
I can see that $M \subseteq M_0$ and therefore $M$ is meager too, but the above step is what I don't get. I can't immediately see why knowing the norms of the $f_n$ is helpful.
 A: I would go a bit different on this. There must be a good way to use the uniform boundedness principle that I cannot see; however here's another approach:
Set for $k\geq1$
$$ M_k=\{\phi\in X:\sup_{n\geq1}|f_n(\phi)|\leq k\}$$
Note that
$$M_k=\bigcap_{n=1}^\infty\{\phi\in X: |f_n(\phi)|\leq k\}$$
so $M_k$ is the intersection of closed sets, hence $M_k$ is closed. Now it is obvious that $M_0=\bigcup_{k=1}^\infty M_k$. We will show that each $M_k$ has empty interior.
Task: Suppose $\phi\in M_d$ for some fixed integer $d\geq1$ and let $\varepsilon>0$. What we need to do is find a function $\psi\in X$ such that $\|\phi-\psi\|_\infty<\varepsilon$ and there exists some $n\geq1$ so that $|f_n(\psi)|>d$.
The main tool: For any integer $k$, let $\omega_k:[0,1]\to\mathbb{R}$ be the function defined as follows: set $\omega_k(\frac{j}{k})=(-1)^j$ for $j=0,\dots,k$ and extend $\omega_k$ linearly in the intervals $[\frac{j}{k},\frac{j+1}{k}]$. Note that, we can choose $n$ so large that $a$ and $a+\frac{1}{n}$ lie in the same interval $[\frac{j_0}{k},\frac{j_0+1}{k}]$. For such $n$, $\omega_k$ satisfies $$|f_n(\omega_k)|=\bigg|\frac{\omega_k(a+\frac{1}{n})-\omega_k(a)}{\frac{1}{n}}\bigg|=2k$$
by its definition (it is linear, so the above quantity is the slope of the line segments that make up its graph; you can compute this by using $|\frac{\omega_k(j/k)-\omega_k((j+1)/k)}{1/k}|=2k$ due to linearity). Note also that, the maximum value that $\omega_k$ attains is $1$, while the minimum is $-1$, so $\|\omega_k\|=1$ for all $k$. We will use these weird "saw-like" functions to create our $\psi$.
We use Weierstrass approximation to obtain a polynomial $p$ such that $\|\phi-p\|_\infty<\varepsilon/2$. Note that polynomials are Lipschitz functions when restricted to bounded intervals (can you see why? it's a fun exercise!), so we have a constant $M_p>0$ such that $|p(x)-p(y)|\leq M_p|x-y|$ for all $x,y\in[0,1]$. This shows that, for any $n\geq1$, $|f_n(p)|\leq M_p$ (we will use that at the final part).
Now let $k=\frac{M_p+d}{\varepsilon}$ and take the function $\omega_k$. Set $\psi:=p+\frac{\varepsilon}{2}\omega_k$. Note that
$$\|\phi-\psi\|_\infty\leq\|\phi-p\|_\infty+\frac{\varepsilon}{2}\|\omega_k\|<\frac{\varepsilon}{2}(1+\|\omega_k\|)=\varepsilon $$
Now as we said when constructing the saw-like functions, we can take a large $n$ so that $|f_n(\omega_k)|=2k$. Then,
$$|f_n(\psi)|=|f_n(p)+\frac{\varepsilon}{2}f_n(\omega_k)|\geq\bigg||f_n(p)|-\frac{\varepsilon}{2}|f_n(\omega_k)|\bigg|\geq$$ $$\geq\frac{\varepsilon}{2}|f_n(\omega_k)|-|f_n(p)|=(M_p+d)-|f_n(p)|\geq(M_p+d)-M_p=d$$
which is what we wanted to show!
