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3

The corrected version is $$u(x)=-\frac{e^{-x}}{2}\int_{-\infty}^{x}f(t)e^{t}dt-\frac{e^{x}}{2}\int_{x}^{\infty}f(t)e^{-t}dt.$$ If $f$ is compactly supported in $\mathbb{R}$, then $u(x) = Ce^{x}$ for large negative $x$, where $C=-\int_{-\infty}^{\infty}f(t)e^{-t}dt$; and $u(x)=De^{-x}$ for large positive $x$, where ...

2

Your convergence statement is false, because Dirichlet's test assumes monotonicity. Indeed $x_n=(-1)^n \frac{1}{n}$ is in $c_0$ but $\sum_{n=1}^\infty (-1)^n x_n = +\infty$.

2

Hint Define a sequence of polynomials by $$P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k$$

2

Let $$X=\{f:\mathbb{R} \to \mathbb{R} \}$$ Then $X$ is a function space, as mentioned in the comments. Note that $M \subset X$. Also note that $M$ is a function space in its own right. What would it mean to call a subset of $X$ open? Well, there's a field of math called topology that studies how one would go about defining this. But if you haven't ...

2

For the second part, take a bounded subset $B$ of $C[a,b]$. We can find a constant $R$ such that $|v(x)|\leqslant R$ for each $x\in [a,b]$ and each $v\in B$. To prove that $M(B)$ is equi-continuous and bounded, use the fact that $g$ is uniformly continuous (and bounded) on $[a,b]\times [-R,R]$.

2

No, the last assumption does not follows from the first two one. To see this consider operators $T_n f = f\left(x^n \right).$

2

From weierstrass approximation theorem, for any $\epsilon > 0$, there exists a polynomial $P_{\epsilon}(x)$ such that $$\int_0^1 |f(x) - P_{\epsilon}(x)| dx \le \epsilon$$ Now, from the condition of this question, $$\int_0^1f(x)P_{\epsilon}(x) dx = 0$$ So, (using fact f is continuous on closed interval) $$\int_0^1f^2(x)dx = \int_0^1f(x)(f(x) - ... 1 A quick remark on your solution of (2): You need to say that \phi = 1 on [0;1] or something like that, as \phi has to be independent of n. I know it's pedantic, but this is how you can loose marks on exams. For (3): you actually want to show that the map$$T(\phi) = \lim_{n \to \infty} \int_{0}^{\frac{1}{n}} n^2f(x)dx - n \delta(\phi)$$is a ... 1 The idea as a whole is correct. If you are not convinced with the equality of the two infimums, you can do as follows: For y \in Y, we have$$y - y' \in Y \implies \|b - (y - y')\| \geq \inf \{ \|b - \tilde{y}\|: \tilde{y} \in Y\} = N(\bar{b}).$$Taking infimum over y \in Y we have$$N(\bar{a}) = \inf \{\|b - (y - y')\|: y \in Y\} \geq N(\bar{b}).$$... 1 [Answer under repair; see comments below] Take p = p_{n}(x) = x^{2n} - x^{2n+1}, so that L(p) = x^{2n}. We note that$$ \|p\| = \int_{-1}^1 |p(t)|\,dt = 2\int_0^1 [t^{2n} - t^{2n+1}]\,dt = 2 \left[ \frac {1}{2n} - \frac{1}{2n+1}\right] = \frac{1}{n(2n+1)} $$On the other hand,$$ \|L(p)\| = \int_{-1}^1 t^{2n}\,dt = \frac{2}{2n+1} $$From there, it ... 1 (1) See the Wikipedia entry on Holder spaces. The Lipschitz norm is the special case \alpha = 1. (2) Yes, this is what it means to prove that f is K-Lipschitz. (3) If f is M-Lipschitz, then f is also K-Lipschitz for any K>M. So if you know f is Lipschitz, then you can assume it has a Lipschitz constant as large as you like; in ... 1 Consider the subspace V_\infty of \mathscr l^2(\mathbb N) where only a finite amount of terms in a series is non-zero. This is an infinite dimensional normed vector space. Define also the subspace V_n where only the first n terms of a series are non-zero. V_n \cong \mathbb R ^n with the standard norm. As such there is a sequence of compacta ... 1 Yes, this is standard. Let T\in A selfadjoint, i.e. with T=T^*. Note first that \phi(T) is real: since T+\|T\|\,\text{id} is positive, we have that$$ \phi(T)+\|T||\in\mathbb R, $$so \phi(T)\in \mathbb R. Now, as -T+\|T\|\,\text{id}\geq0, we get -\phi(T)+\|T\|\geq0, so$$\phi(T)\leq\|T\|. Since $-T$ is also selfadjoint, we can also get ...

1

Counterexample: $f(x)=\frac{1}{x}\in L^2([1,+\infty))$, but it does not belong to $L^1([1,+\infty))$

1

For $1\leq p<q<\infty ,\quad$ $f(x)=x^{-1/q}$ belongs to $L^p(0,1)$ but not to $L^q(0,1)$.

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