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Hint: A compact space is separable. Write $X=U_nB(0,n)$ $K(B(0,n)$ is separable since it is relatively compact, $\bigcup_nK(B(0,n))=K(X)$ is separable since it is the union of separable spaces. For your second question let $(v_n\neq 0)$ be a dense family in $Y$, write $w_n={{v_n}\over{\|v_n\|}}$. Consider $K:l^1\rightarrow Y$ defined by $K(e_i)={w_i\over i}$...

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Alternative answer:your operator induces a continuous bijection between quotient space and image subspace. By the open mapping theorem if it were compact then the unit ball in both spaces were compact, thus finite dimensional by Riesz

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First, note that $Z:=Im(K)$ is a closed subspace of a Banach space and thus, itself a Banach space. Thus, $K: X\to Z$ is onto. By the open mapping theorem, $K$ is open and hence, $K$ is mapping open sets to open sets. Now, assume that $K$ is compact and take the image of the open unit ball $C:=K(B_X^\circ)$ which is open in $Z$ and relatively compact in $Y$...

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In fact, $A$ can't be a compact operator (if $X$ is infinite dimensional). This condition is sometimes referred to as "$A$ is bounded away from $0$". To show that it can't be compact, it suffices to show that there is a bounded sequence whose image has no convergent subsequence. Note (by the usual application of Riesz's lemma) that there exists a bounded ...

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If $n > m$, then for $x\in [m+1/2,m+1]$, $$g_n(x) = 0 \text{ but } g_m(x) = x-m \geq 1/2$$ Hence, $$\|g_n - g_m\| \geq 1/2 \quad\forall n\neq m$$ so it can't have a Cauchy subsequence.

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Hint: Fix any $n\in\mathbb{N}$. Then, show that for any $m>n+1$, we have $\|g_n-g_m\|_{\infty} \geq 1$.

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Hint: $g_n$ is weak-$*$ convergent.

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Note that $L$ is invertible with $L^{-1} (f) = (1+x) f$. Thus the composition $L\circ L^{-1}$ is the identity. If $L$ is compact, then so is $I$, which is not true.

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Note that the range of $K$ is in a finite dimensional subspace $$\text{span} \{x, \cos x\},$$ thus $K$ is a compact operator and has only point spectrum ($0$ is a eigenvalue too). To look for eigenfunction, we need only restrict ourselve to the above subspace. Note that $$K (ax+b \cos x) = b\pi x + a\frac{2\pi^3}{3} \cos x,$$ This implies that $$... 2 Note that$$ \left (\sum_j a_{ij} x_j \right)^2 = \lvert \langle (a_{ij})_j, (x_j)_j \rangle \rvert^2 \leq \lVert (a_{ij})_j \rVert_1^2 \rVert x\lVert_1^2 So, if x is summable, you know that \lVert x \rVert_1 < \infty and you can get rid of the norm of x in the estimate. By \langle \cdot,\cdot \rangle I mean the inner product in \ell^1. 0 You have to understand what A looks like. It acts as an infinite matrix (a_{ij})_{i,j=1}^{\infty} on your elements of \ell^2. Since \ell^2 is a Hilbert space you know that A is compact if it is a limit of finite-rank operators. You can show that A is the limit of P_nAP_n where P_n is the projection on the span of the first n standard basis ... 2 For f\in L^2 and 0 \le r \le s \le 1, \begin{align} |(Af)(r)-(Af)(s)| & \le \int_{r}^{s}|f(x)|dx \\ & \le \left(\int_{r}^{s}|f(x)|^2dx\right)^{1/2}\left(\int_{r}^{s}dx\right)^{1/2} \\ & = \|f\|\sqrt{s-r}. \end{align} Therefore the imagine of a bounded subset of L^2 is uniformly bounded and equicontinuous, ... 4 For Questions 1, take \{f_n\} be uniformly bounded in L^2, consider the class \{Af_n\}. check 1.\{Af_n\} is uniformly bounded in L^2. check 2.\{DAf_n\} is well-define and uniformly bounded in L^2, where D is Differential operator. pf of check 1: This follows by Holder, |\int_0^t f(s) ds| \leq \sqrt{t }\|f_n\|_2\leq \|f_n\|_2, so \|... 3 For determining the adjoint : You have that\langle Af, g \rangle = \int_0^1 g(t) \int_0^t f(s) ds dt $$And this is equal (by Fubini) to$$ = \int_0^1 f(t) \int_t^1 g(s) ds dt = \langle f, Bg \rangle$$with Bg(t) = \int_t^1 g(s) ds And as this is true for every f and g, you have that A^* = B 1 It is clear that 0 is in the spectrum of K, since K is compact. Let us address if it is an eigenvalue: if$$ Kf=0,$$this means we have$$\tag{1} 0=t\int_t^1f(s)\,ds+\int_0^ts\,f(s)\,ds. $$Differentiating (via Lebesgue's Differentiation Theorem),$$\tag{2} 0=\int_t^1 f(s)\,ds-tf(t)+tf(t)=\int_t^1f(s)\,ds,\ \ \ \text{a.e.}. $$Then, for any v,t\in[0,... 1 To your first question, let$$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $$Then$$ p\vee q=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ p+q=\begin{bmatrix}3/2&1/2\\1/2&1/2\end{bmatrix}, $$so$$ p+q-p\vee q=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}, $$which is ... 1 Note that the two equalities A=AA^*A and A^*=A^*AA^* are the same, since you can obtain one from the other by taking adjoints. Assume first that A=AA^*A. By multiplying by A^* on the left, we get$$ A^*A=(A^*A)^2. It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...

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No, it's not a coincidence. The definition of "$\lim (T-T_k)=0$" is "$\lim||T-T_k||=0$". No, you can't do the same for any operator. You can't show that $||T-T_k||\to0$. (Let $Tx=x$. What is $||T-T_k||$?)

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