5

Let $f \ne 0$ be an unbounded linear functional on $X$. Then there is $u \in X$ such that $f(u)=1.$ Now define $P:X \to X$ by $$P(x):=f(x)u$$ $P$ will do the job.


4

Write $p(x)=\displaystyle\sum_{k=0}^{N}a_{k}x^{k}$, then \begin{align*} &(1-x)\sum_{n=0}^{\infty}x^{n}p(x^{n})\\ &=(1-x)\sum_{n=0}^{\infty}\sum_{k=0}^{N}a_{k}x^{n}(x^{n})^{k}\\ &=(1-x)\sum_{k=0}^{N}a_{k}\sum_{n=0}^{\infty}x^{n(k+1)}\\ &=(1-x)\sum_{k=0}^{N}a_{k}\dfrac{1}{1-x^{k+1}}\\ &=\sum_{k=0}^{N}a_{k}\dfrac{1}{1+x+\cdots+x^{k}}\\ &\...


3

You take $f=1$. Then $\lVert f\rVert_\infty=1$ and $\bigl\lvert T(f)\bigr\rvert=\mu\bigl([0,1]\bigr)$.


3

Why should you be able to conclude that $||x||_1=||x||_2$? This is just not true. Consider $X=\mathbb{R}^n$ with $||\cdot||_1=||\cdot||_{L^1}$ and $||\cdot||_2=||\cdot||_{L^\infty}$. Then these norms are equivalent (in particular, both identity maps are continuous) but not identically equal.


2

Notice that $|f|^p\leq 1_{\{|f|\leq 1\}}+|f|^q1_{\{|f|> 1\}}$ for $0<p\leq q$. The function on the right is integrable since $$ \int_X 1_{\{|f|\leq 1\}}+|f|^q1_{\{|f|> 1\}} \leq \mu(X)+\int_X |f|^q<\infty. $$ Now we can apply DCT since $|f|^p\leq 1_{\{|f|\leq 1\}}+|f|^q1_{\{|f|> 1\}}$. We have $$ \lim_{p\to 0^+}\int_X |f|^p=\int_X \lim_{p\to ...


2

Not every subset of $\mathbb R^{2}$ is a measurable set. If $A$ is a non-measurable set then define $f(x)=1$ for $ x \in A$ and $f(x)=0$ for $ x \in A^{c}$. Then $f$ is not measurable. If $E$ is not measurable in $\mathbb R$ then $E \times \mathbb R$ is not measurable in $\mathbb R^{2}$. For an example of such a set $E$ see https://www.encyclopediaofmath....


2

Let $M$ be the closed subspace spanned by $(e_n)$. Then $(e_n)$ is an orthonormal basis for $M$ and every element $x$ of $M$ is of the type $\sum \langle x, e_j \rangle e_j$. It follows that $x \in V$. On the other it is also obvious that $x \in V$ implies $x \in M$. Hence $V$ is the closed subspace spanned by $(e_n)$.


2

Fix a sequence $y_n \in S_X$ such that $y_n \to y$. First, by the reverse triangle inequality, we have that $$\|x + \lambda y_n\| - \|x+ \lambda y\| \leq |\lambda| \|y_n - y\|.$$ For fixed $\varepsilon > 0$ pick $\lambda_0$ such that $\|x + \lambda_0 y\| \leq \|x + \langle y \rangle \| + \frac{\varepsilon}{2}$. Then we get that $$\|x + \langle y_n \...


2

Here is a proof for $p>1$. We want to show the existence of $c>0$ such that $$ \|u \|_{W^{n,p}} \le c ( \|u \|_{L^p} + \|D^nu\|_{L^p}) \quad \forall u\in W^{n,p}(0,1). $$ Assume this is not true. Then for every $k$ there is $u_k$ such that $$ \|u_k\|_{W^{n,p}} > k ( \|u_k\|_{L^p} + \|D^nu_k\|_{L^p}). $$ Clearly, $u_k\ne0$. We can rescale the $u_k$'s ...


1

Consider any decreasing sequence $(p_{n})$ to zero. Write \begin{align*} \int|f|^{p_{n}}=\int_{0<|f|\leq 1}|f|^{p_{n}}+\int_{|f|>1}|f|^{p_{n}}. \end{align*} The second integral is monotone decreasing, and $\displaystyle\int_{|f|>1}|f|^{p_{1}}$ is integrable (we choose $p_{1}$ small enough such that $p_{1}<q$), so it converges to $\displaystyle\...


1

Keep in mind that functional derivatives are taken by varying the function, not its argument. So unless the functional itself involves derivatives, and integration by parts is involved in transforming the result (as in the Euler-Lagrange equations, for example) there is no reason for derivatives of the function to appear. Consider an even better case with ...


1

Take a non closed vector subspace $E$ of codimension $1$ and $u$ not in $E$, for every $x\in X$, there exists unique real $c(x)$ and $P(x)$ such that $x=P(x)+c(x)u$, $P^2(x)=P(x)$ and its not bounded. https://mathoverflow.net/questions/30868/subspaces-of-finite-codimension-in-banach-spaces


1

Let $f$ be an unbounded linear functional and $P(x)=f(x) x_0$ where $x_0$ is a fixed vector with $f(x_0)=1$. Then $P$ is not continuous and $P^{2}=P$.


1

Note that this is a statement about unbounded operators. Those are not necessarily defined everywhere. The statement, as a matter of fact, is completely trival, since the graph of $A$ is mapped to the graph of $A^{-1}$ under the homeomorphism between $X\oplus Y$ and $Y\oplus X$ which flips the coordinates. Note also that if $Dom(A)\subsetneq X$, then $A$ ...


1

Meanwhile someone could add something about the references, let me tell you how the Frechét technique is used to get the famous Euler-Lagrange condition with a classical example. Let us suppose the you have a integral functional like $$A=\int_I{\cal L}(x,y,y')dx$$ where $y=y(x)$. Now to mimic Frechét procedure we take $$A_{\varepsilon}=\int_I{\cal L}(x,y+...


1

First note that if a sequence $\{f_n\}_{n\in\mathbb{N}}\subset AC([a,b])$ is a cauchy sequence in the sensee of your norm, then $\{f_n\}_{m\in\mathbb{N}}$ and $\{f_n^{'}\}_{m\in\mathbb{N}}$ are cauchy sequences in the sense of the $L^1$ norm, so you can argue the existence of functions $f$ and $g$ such that $f_n\rightarrow f$ and $f_n^{'}\rightarrow g$ in ...


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