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16h
revised How can I solve like this exercise
edited tags
2d
comment Is the following statement true on $L^0$ spaces?
This answer is not true. If $f(x) = x$ would suffice, the conclusion would be that $X = Y$ whenever $E(X) = E(Y)$. This is obviously wrong.
Apr
11
answered Self-adjointness under relatively bounded perturbation
Apr
10
answered Approximating $u \in H^1$ s.t. $u(T)=0$ with $u_n \in H^1_0$ in the gradient norm?
Mar
13
answered separation of convex cone
Mar
13
comment sobolev spaces integral estimation
But you have the strong convergence of the gradient on $\tilde\Omega$.
Mar
9
comment If $f:\mathbb C \to \mathbb C$ is continuous at a point $z_0$, then show that $\overline {f(\bar z)}$ is also continuous at $z_0$.
I think so. You can take any function $f$ which has exactly one (non-real) point of discontinuity, which you call $\bar z_0$.
Mar
9
comment If $f:\mathbb C \to \mathbb C$ is continuous at a point $z_0$, then show that $\overline {f(\bar z)}$ is also continuous at $z_0$.
Should it be "is also continuous at $\bar z_0$"? Otherwise it should be false.
Mar
8
comment On the Definition of Gateaux Derivative
No, I don't think so.
Mar
7
comment On the Definition of Gateaux Derivative
Yes, I know weak derivatives.
Mar
6
comment On the Definition of Gateaux Derivative
Derivative with increment $h-f$ ;)
Mar
6
comment On the Definition of Gateaux Derivative
If you replace $h$ by $h - f$ in the first definition, you arrive at the second. Hence, they define different things. The typical definition is the first one.
Mar
6
comment To prove (X*)**= (X**)*
Typically, one defines $X^{**} = (X^*)^*$. But then, $(X^*)^{**} = ((X^*)^*)^* = (X^{**})^*$.
Mar
4
comment Continuous iff composition with every linear functional is continuous
Using the last inequality and taking the sup over all $y'$ with norm at most $1$, we get $\|T(x)\| \le C \, \|x\|$. But this yields $\|T\| \le C$.
Feb
21
comment Proving finite dimensional normed linear space is complete , without using equivalence of norms on finite dimensional vector spaces
Another possibility would be to use the precompactness of bounded sets. So your Cauchy-sequence has a convergent subsequence and, hence, converges.
Feb
19
comment Square root of the operator $T$
What have you tried?
Feb
19
revised Square root of the operator $T$
Beautified question.
Feb
18
accepted Reference: Continuity of Eigenvectors
Feb
17
comment Preimage of Legendre-Fenchel transform
In the reflexive case, there are some assertions like "the biconjugate equals the convex lsc envelope", that is the largest, convex, lsc function below your original function. Then, you get all functions $g$, whose convex envelope is the conjugate of $f$. Maybe you can argue similar in the non-reflexive case.
Feb
17
comment Preimage of Legendre-Fenchel transform
Are you interested in the case that $X$ is not reflexive? I think in the reflexive case one can argue by using the convex envelope.