Let $X$ and $Y$ be normed spaces and $T: D(T)\rightarrow Y$ a linear operator, where $D(T)\subset X$. The operator $T$ is said to be unbounded if there exists a sequence ${x_n}\subset D(T)$ s.t. $$| Tx_n| \geq n| x_n| $$
Questions tagged [unbounded-operators]
404 questions
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Bounded kernel operator
I have to verify that this operator in $L^2(\Bbb R^3): (Gf)(x)=\int dy f(y) \frac{e^{-|x-y|}}{|x-y|} $ is bounded, using the following rule for the integral kernel: $ sup_x\int dy|a(x,y)|< \infty$ and
$sup_y\int dx|a(x,y)|< \infty$, where $|a(x,y)|$…
user930969
- 21
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Define a unbounded functional on C00 ⊂ l 2 . Justify your answer
I know that Not possessing both an upper and a lower bound. So for all positive real values V there is a value of the independent variable x for which |f(x)|>V. For example, f (x)=x 2 is unbounded because f (x)≥0 but f(x) → ∞ as x → ±∞, i.e. it is…
91Hana01
- 11
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When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its maximal domain?
Suppose we have a closable and densely defined operator $A$ with a domain $dom(A)$ which is a subspace of a Hilbert space $\mathcal{H}$.
Let $\mathcal{H}$ have an orthonormal basis $\{e_n\}_{n=1}^\infty$.
So the operator $A$ can be viewed as an…
apyshkin
- 11
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Positive symmetric densely defined operator with dense range: essentially self-adjoint?
Suppose $A$ is densely defined, symmetric ($\langle Ax,y \rangle = \langle x,Ay \rangle$ for $x,y \in \textrm{dom}(A)$), and semi-strictly positive ($\langle Ax, x \rangle > 0$ for all $0 \neq x \in \textrm{dom}(A)$). If $A$ has dense range, is $A$…
marbled
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Show that the set is not bounded - formal proof - $D = \{(x,y)\in R^2|y=\frac 1 x\}$
is there any way to prove with epsilon \ other way that the set is unbounded?
$$D=\{(x,y)\in R^2\mid xy=1\} = \{(x,y)\in R^2\mid y=\frac 1 x\}$$
Here is a proof my exerciser at class did ( which is not enough proof for me, If there is way to prove…
user1112370
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uniform boundedness of derivativative
$X={f\in C(0,1): f^{\prime \prime }(t)<0}$.
$T:X\mapsto X$ is defined by:
$Tf(t) = at+ (1-t)\int_{0}^{t}sf(s)ds + t\int_{t}^{1}(1-s)f(s)ds$
and it is known that $\int_{0}^{1}s(1-s)f(s)ds < \infty$.
So that I have:
$|\frac{d}{dt}Tf(t)| < a +…
