- State the Banach-Steinhaus Theorem about convergence of a sequence of linear operators in normed spaces.
- Let $X=C([0,1])$ be the Banach space of all functions continuous on $[0,1]$ with the standard norm $\|⋅\|_{∞}$. Let $(α_n)$ be a given sequence of positive numbers converging to zero as $n → ∞$. Define $A_n: X → X$ so that $A_n x(t)=x(t^{1+α_n})$. Show that $A_n x → x$ in $X$ for any $x ∈ X$.

- Let $X$ and $Y$ be Banach spaces and consider a sequence $T_n ∈$ $ℬ(X, Y)$. The following statements are equivalent:
- There exists $T ∈ ℬ(X, Y)$ such that, for every $x ∈ X, T_n x → T x$ as $n → ∞$.
- For each $x ∈ X$, the sequence $(T_n x)$ is convergent.
- There is a constant $M$ and a dense subset $Z$ of $X$ such that $\|T_n\| ⩽ M$ and the sequence $(T_n z)$ is convergent for each $z ∈ Z$.

- Since the function $t ↦ t^{1+α_n}$ is one-to-one from $[0,1]$ onto $[0,1]$, therefore $\|A_nx\|_{∞}=\|x\|_{∞}$. Hence, $\|A_n\| ⩽ 1$.

Now, let $x$ be a continuously differentiable function on $[0,1]$. By mean value inequality, we have \[ |A_n x(t)-x(t)| ⩽\|x'\|_{∞}(t-t^{1+α_n}) \] for all $t ∈[0,1]$. Consider $g(t)=t-t^{1+α_n}$ and find $\|g\|_{∞}$. Since $g(0)=g(1)=0$, we need to find $0<t_0<1$ such that $g'(t_0)=0$. It is clear that $t_0=(\frac{1}{1+α_n})^{\frac{1}{α_n}}$ and $(\frac{1}{1+α_n})^{\frac{1}{α_n}} →\frac1e$ so$$g(t_0)=\frac{α_n}{1+α_n}(\frac{1}{1+α_n})^{\frac{1}{α_n}} → 0\text{ as }n → ∞$$So, $A_n x → x$ in $X$ for continuously differentiable function. Such functions are dense in $X$ and the result follows from Banach-Steinhaus theorem.

- State the Open Mapping Theorem.
- Let $X$ be a Banach space, $Y$ be a normed space, and $T ∈ ℬ(X, Y)$. Denote by $B_X$ the unit ball of $X$ centred at the origin. Show that if the interior of $\overline{T(B_X)}$ is not empty, then the interior of $T(B_X)$ is not empty either.
- Let $X$ and $Y$ be Banach spaces, and $T ∈ ℬ(X, Y)$. It is said that a subset of $Y$ is of the
*first category*in $Y$ if it is a countable union of nowhere dense sets in $Y$. Prove that either $T(X)$ is closed or of the first category in $Y$. All theorems used should be clearly stated. - Let $L^1(0,1)$ be the Lebesgue space of equivalence classes of integrable functions on $(0,1)$ equipped with \[ \|x\|=∫_0^1|x(t)|\mathrm d t . \] Show that the subspace of $L^1(0,1)$ that consists of all equivalence classes of integrable functions containing a function from $C([0,1])$ is of the first category in $L^1(0,1)$.

- Let $T ∈ ℬ(X, Y)$, where $X$ and $Y$ are Banach space. Let $T$ be onto. Then $T$ is open.
- Now, let us assume that $T: X → Y$ is linear and continuous, $X$ is Banach, $Y$ is normed, and $\overline{T(B_X)} ⊃ B_Y(ε)$ for some positive $ε$. We are going to prove that there is $δ>0$ such that $T(B_X) ⊃ B_Y(δ)$. Error! Click to view log. Let $y ∈ \overline{T(B_X)}$. Then we can find $y_1 ∈ T(B_X)$ such that $\|y-y_1\|_Y<ε / 2$. Since $y-y_1 ∈ B_Y(ε / 2), y-y_1 ∈ \overline{T(\frac{1}{2} B_X)}$ and thus we can find $y_2 ∈ T(\frac{1}{2} B_X)$ such that $\|y-y_1-y_2\|_Y<ε / 4$. Proceeding in this way, we find sequences $y_n$ and $x_n$ with the following properties: \[ \|y-y_1-y_2-…-y_i\|_Y<ε / 2^i, y_i=T x_i, x_i ∈ \frac{1}{2^{i-1}} B_X, i=1,2, …, n \] for all $n$. Since $X$ is Banach and $\|x_n\|_X ⩽ \frac{1}{2^{n-1}}$, the sequence $z_n=\sum_{i=1}^n x_i$ converges to $z ∈ X$. Moreover, $\|z_n\|_X ⩽ \sum_{i=1}^n \frac{1}{2^{i-1}}<2$. Hence, we have $\|y-T z_n\|_Y<ε / 2^n$ and passing to the limit, we show that $y=T z$ with $z ∈ 2B_X$. So, one can take $δ=ε/2$.
- There are two options: either the interior of $\overline{T(B_X)}$ is empty or it is non-empty. In the first case, we have $T X=⋃_n T(n B_X)$, where $T(n B_X)$ is a nowhere dense set.

In the second case, we can use (ii) and conclude that the interior of $T(B_X)$ is not empty which, by linearity of $T$, implies $T X=Y$, so $TX$ is closed. - $X=C([0,1])$ equipped with sup norm and $Y=L^1(0,1)$ equipped with $L^1$ norm are Banach spaces. Let $π: X → Y$ be the quotient map (for $x ∈ C([0,1])$, $π x$ is the equivalence class in $L^1(0,1)$ containing $x$). It is continuous as $\|π x\|_Y=\int_0^1|x|⩽\sup|x|=\|x\|_X$ but $πX$ is not closed since continuous functions are dense in $L^1(0,1)$. The result follows from (iii).