Hot answers tagged

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 $L^q(0,1)$.



Only top voted, non community-wiki answers of a minimum length are eligible