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Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite For $f in L^2([0, 1]) $ and $x in [0, 1]$ we have the following $$ A(f)(x) = int_0^1 (x - y) f(y)dy $$ In this particular exercise I have to calculate the operator norm $| A |$. We know that from a theorem that if $A$ is compact and self-adjoint then either $|A|$ or $-| A |$ is an eigenvalue to the operator $A$. My attempt: We know that $A$ is indeed compact, but it is not self-adjoint. However, by some inspection, we have that $iA$ is self-adjoint (and compact). So we just have to calculate $|iA| = |A| = |lambda|$ where $lambda$ is an eigenvalue satisfying $A(f)(x) = lambda f(x)$, for some eigenfunction $f$. Now setting $f(x) = a + bx$ we get by calculation that $$lambda(a + bx) = int_0^1 (x - y) (a + by) dy = -frac a 2 - frac b 3 + (a + frac b 2)x.$$ But from this I'm not sure how to solve for $lambda$. Maybe there is an another way to solve this? functional-analysis share | cite | improve this questi

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite I have a continuous process for which I want to compute the probability that it did not fail until time $T$. I tried the reliability / survival function: $$ beginalign R(T) = exp(-int_t=0^T h(t) dt) endalign $$ I do not have a failure rate $h(t)$, but a failure probability given by a continuous but not necessarily monotonic function $p_fail(t)$, which gives the probability that the process fails at any time $t$ (with $t in R, t ge 0$). Here, $p_fail(t)$ does not have a unit, which messes up the whole calculation. Therefore, I think my approach is wrong. How can I calculate the reliability of the process? To illustrate my problem: Consider the simple function $p_1,fail(t [s]) = frac0.013600s cdot t$, for $t$ in seconds. This (in my opinion) is equal to $p_2,fail(t [h]) = frac0.011h cdot t$, for $t$ in hours. With $T = 1h = 3600s$ the integration gives different results, depending on the unit of time: beginarray c h

Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Lemma 3.43 Suppose $B_m$ are elements of $G$ and that $B_m to I$. Let $Y_m = log B_m$, which is defined for all sufficiently large $m$. Suppose that $Y_m$ is nonzero for all $m$ and that $Y_m/|Y_m| to Y in operatornameM_n(mathbbC)$. Then $Y$ is in $mathfrakg$. Proof. For any $t in mathbbR$, we have $(t/|Y_m|)Y_m to t Y$. Note that since $B_m to I$, we have $|Y_m| to 0$. Thus, we can find integers $k_m$ such that $k_m |Y_m| to t$. We have, then, $$ e^k_m Y_m = exp left[ (k_m |Y_m|) fracY_mY_m right] to e^tY. $$ (Original image here.) The group $G$ is a Lie group of matrices. This was taken from Brian Hall. It’s reallly just analysis question. But I’m not entirely sure why the $k_m$’s exists real-analysis share | cite | improve this question edited Aug 23 at 10:37 Jendrik Stelzner 7,572 2 10 37 asked Aug 22 at 4:55 Hawk 5,273 9 36 99 add