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I am a little late to the question but wanted to add a low-tech answer which somehow complements JDH's answer: The operators $\frac{d}{dx}$ and $\frac{d}{dt}$ are as distinguishable as $f(x)$ and …

If you do not need an explicit form you could start with $$ f_\epsilon(s) = \begin{cases} \frac{s}{\|s\|}, & \|s\|\geq \epsilon\\ 0, & \|s\|<\epsilon \end{cases} $$ and then use a narrow mollifier $\p …

While @Zander is right about the fact that "these functions cannot make a basis for the functions space" there are still a lot of families of translates of "bell shaped functions" which for a dense se …

Probably you refer to some theorem in "Mathematics of Computerized Tomography" by Frank Natterer (e.g. Theorem 4.2)? Then you are assuming that the domain in $\mathcal{S}$ and if I remember correctly, …

The optimization problem for $f_3$ is $$ \min_g \int_{-1}^1 |g''(x)|^2dx \quad \text{s.t.}\quad \int_{-1}^1 g(x)dx = 0,\ \int_{-1}^1 x g(x)dx = 0,\ \int_{-1}^1 |g(x)|^2dx = 1. $$ Using Lagrange Multip …

Consider real polynomials on the interval $I=[-1,1]$. It is easy to see that the smallest degree for a non-negative polynomial with given zeros $x_1,\dots,x_s\in I^\circ$ is $n=2s$ (e.g. $P(x) = \prod …

That's a standard result in convex optimization. For example Theorem 2.1.5 in Nesterov's "Introductory Lectures on Convex Optimization" states that the following are equivalent: $f$ is $C^1$, convex …

Edit: Note that I understood the question for $L^1([0,1]^3)$, c.f. Robert Israels comment. First, the term "basis" for general Banach spaces, especially ugly spaces such as $L^1$, can be complicated. …

Another construction that works is the following: Take any convex $A\subset\mathbb{R}^n$ which contains a neighborhood of the origin. Then the associated Minkowski functional $$ \sigma_A(x) = \inf\{\l …

Here is my two cents: In (unconstrained continuous) optimization you want to find minima of functions and in the differentiable case these have vanishing gradient. (In the convex case, vanishing gradi …

Although this question sounds quite innocent, a systematic treatment of higher order derivatives of implicit functions is quite involved. On a second thought, this is no surprise if you think about ho …

When I was introduced to measure theory, the professor chose to use the Choquet integral to obtain the Lebesgue integral. An this uses the "good old" Riemann integral to integrate the pseudo-inverse o …

Probably the Fenchel inequality counts? $\newcommand{\RR}{\mathbb{R}}$ For a function $f:X\to\RR\cup\{\infty\}$ on a vector space $X$ and it's convex conjugate $f^*(x^*) = \sup_{x\in X}\langle x^*,x\ …