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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. Apr11 answered Self-adjointness under relatively bounded perturbation Apr10 answered Approximating $u \in H^1$ s.t. $u(T)=0$ with $u_n \in H^1_0$ in the gradient norm? Mar13 answered separation of convex cone Mar13 comment sobolev spaces integral estimation But you have the strong convergence of the gradient on $\tilde\Omega$. Mar9 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 $\bar z_0$. Mar9 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. Mar8 comment On the Definition of Gateaux Derivative No, I don't think so. Mar7 comment On the Definition of Gateaux Derivative Yes, I know weak derivatives. Mar6 comment On the Definition of Gateaux Derivative Derivative with increment $h-f$ ;) Mar6 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. Mar6 comment To prove (X*)**= (X**)* Typically, one defines $X^{**} = (X^*)^*$. But then, $(X^*)^{**} = ((X^*)^*)^* = (X^{**})^*$. Mar4 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 $\|T\| \le C$. Feb21 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. Feb19 comment Square root of the operator $T$ What have you tried? Feb19 revised Square root of the operator $T$ Beautified question. Feb18 accepted Reference: Continuity of Eigenvectors Feb17 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. Feb17 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.