Operator of differentiation Okay I need to know once and for all how to properly show that $L$ is closed and not continuous, $L_n$ is bounded for all $n \in \mathbb{N}$, $L_n \to L$ pointwise when $n \to \infty$, $L_n$ doesn't converge to $L$ with respect to $\| \cdot \|_{\mathrm{op}}$. I have this example:
Consider for $X = (C^0([0,1]), \| \cdot \|_\infty)$ the subspace $D = C^1([0,1]) \subset X$. The operators $L, L_n \colon D \to X$ are defined as $Lf := f'$ and
$$
  (L_n f)(t)
  :=
  \begin{cases}
    \dfrac{f(t)-f(t-\frac{1}{n})}{\frac{1}{n}}
    & \text{for $t \in [\frac{1}{n}, 1]$},
    \\[1em] % some extre spacing between the two cases
    \dfrac{f(\frac{1}{n}) - f(0)}{\frac{1}{n}}
    & \text{else}.
  \end{cases}
$$
for $n \in \mathbb{N}$.
I know this is a big ask, I just want to know, how do you prove these things, so if anyone has any ideas, hints, anything I will be grateful.
 A: The first claim is that the operators $L_n$ are all bounded in operator norm. This is relatively easy to prove: the difference of two bounded operators is bounded, and dividing by a constant $\frac 1 n$ doesn't change the operator being bounded.
The next claim is that $L_n \to L$ pointwise. This is also pretty simple. Pointwise convergence just requires that if we fix $f$ and $t$ then we get $L_n(f)(t) \to L(f)(t)$. That's true basically by definition of derivatives, since $f'(t) = \lim_{n \to \infty} \frac{f(t) - f(t-\frac 1 n)}{1/n}$.
Now for the tricky part: showing that $L_n$ doesn't converge to $L$. To show this, define functions $f_n \in D$ such that

*

*$f_n(t) = 0$ for $0 \le t \le 1-\frac 1 {2n}$

*$f_n(1) = 1$

*In the interval $(1 - \frac{1}{2n}, 1)$, I require $0 \le f_n(t) \le 1$. Besides that, just choose any way to connect the two sides while making $f_n \in C^1$. For example, we could use $f_n(t) = 4n^2 \left(t-1+\frac{1}{2n}\right)^2$. The exact definition won't matter though.

By definition of the operator norm, $\|L - L_n\|_{op} \ge \frac{\|L(f_n) - L_n(f_n)\|_\infty}{\|f_n\|_\infty}.$ I'll show that the RHS does not converge to $0$ as $n \to \infty$, thus proving that $L_n \not \to L$.
First, the denominator: $\|f_n\|_\infty = 1$ since that's the largest absolute value $f_n$ takes.
Next, the numerator.

*

*Evaluating $\|L_n(f_n)\|_\infty$ comes down to finding $t$ such that $\left|f_n(t) - f_n(t - \frac 1 n)\right|$ is as big as possible. The way I constructed $f_n$ means that the best choice is $t=1$, giving a norm of $\|L_n(f_n)\|_\infty = n \left( f_n(1) - f_n(1 - \frac 1 n)\right) = n \left(1\right) = n$.

*Evaluating $\|L(f_n)\|_\infty$ means finding the largest absolute value of $f_n'$. I  know $f_n(1-\frac 1 {2n}) = 0$ and $f_n(1) = 1$, so by Mean Value Theorem I know there's some $t^* \in \left(1 - \frac 1 n, 1\right)$ satisfying $f_n'(t^*) \ge \frac{1}{1/2n} = 2n$. In other words, $\|L(f_n)\|_\infty \ge 2n$.

*Putting those together: $\|L(f_n) - L_n(f_n)\|_\infty \ge \|L(f_n)\|_\infty - \|L_n(f_n)\|_\infty \ge 2n - n = n$.

Now combine everything:
$$
\begin{align*}
\|L-L_n\|_{op} &\ge \frac{\|L(f_n) - L_n(f_n)\|_\infty}{\|f_n\|_\infty} \\
&\ge \frac{n}{1} = n
\end{align*}
$$
In particular, we do NOT have $\|L-L_n\|_{op} \to 0$ as $n \to \infty$, so $L_n \not \to L$ in operator-norm.

An obvious followup question would be: How did I think of choosing those particular functions $f_n$? My answer is: to disprove the convergence, I need to choose functions such that $\|L_n(f_n)\|_\infty$ is very different from $\|L(f_n)\|_\infty$. The operators $L_n$ are good approximations to $L$ if the function values' swings happen across larger ranges (length at least $\frac 1 n$), so I tried choosing functions $f_n$ that change very suddenly across an interval with size significantly smaller than $\frac 1 n$.

Edit: Actually, $L$ is not a continuous (aka bounded) operator at all. Taking this route could have saved me a bunch of trouble in the proof above. Let me explain.
To show $L$ is not a bounded operator: Choose functions $f_n \in D$ such that $\|f_n(t)\| \le 1$ for all $t \in [0,1]$ but $f_n'(t) \ge n$ for at least one value $t$. (Finding such functions is easy; the functions I used in my proof above work fine.) Then $\frac{\|L(f_n)\|_\infty}{\|f_n\|_\infty} \ge n$, so if $\|L\|_{op}$ existed then it would need to be $\ge n$ for all integers $n$. We conclude $L$ is unbounded.
To show it's not possible for bounded operators to converge in op-norm to an unbounded operator: We have $L_n$ bounded and $L$ unbounded. If $L_n - L$ were bounded, then we'd have $\|L\|_{op} \le \|L_n\|_{op} + \|L-L_n\|_{op} < \infty$ using the triangle inequality, so $L$ would be bounded which is a contradiction. Therefore $L_n-L$ is unbounded for all $n$, so we certainly don't have $\|L_n - L\|_{op} \to 0$ as $n \to \infty$.
So $L$ is not a continuous operator and $L_n$ don't converge to $L$ in op-norm even though $L_n \to L$ pointwise.
Finally, we show $L$ is closed. Let $g_n$ be some sequence of functions in $D$ such that $g_n \to g$ and $g_n' \to h$ for some $g, h \in X$. We need to show $g \in D$ and $g' = h$. To show this, we note that convergence in $\infty$-norm is the same as uniform convergence of functions. If the original functions converge to $g$ and the derivatives converge uniformly to $h$, then a standard theorem says $g$ is differentiable and $g' = h$, which is what we needed to prove. If you want to read about that theorem, you could check Theorem 6.3.1 here or this section on Wikipedia.
